Zipless Fun
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For solving: https://sudokupad.app/evwy8a13gv

1. In row 3, the 15-cages consist of the digits 6789. That leaves only two possibilities for the 9-cage in box 2, both of which contain 3. R3C3 and R3C7 are therefore either 15 or 24, whichever pair isn't used in the 9-cage.

2. In box 1, the cages together add up to 24, so the uncaged cells add up to 21. Two-thirds of the pairs on the zipper line have one digit in a 15-cage and the other digit in a 9-cage.
2a. To think about possible combinations, it might help to think about the caged digit that isn't part of a zipper pair, R1C3. Since the total of the five caged digits is even, and the four zipped digits combined are even, so too is the leftover. That already reduces the candidates for the 9-cage to either 126 or 234, with R1C3 being either 2, 4, or 6.
2b. Counting down, that makes the zipper value either 9 (if the 15-cage is 78, the 9-cage is 126, and the un-caged pair is 45), 10 (if the 15-cage is 78, the 9-cage is 234, and the uncaged pair is 19), or 11 (if the 15-cage is 78, the 9-cage is 234, and the uncaged pair is 56). In all of those cases, the 15-cage is a 78 pair.

3. Box 3 differs from box 1 only in cage values. And the fact that we know the 15-cage is a 69 pair from the start. Together the cages add up to 26, so the uncaged cells add up to 19.
3a. Again thinking about combinations by focusing on the unzipped caged digit in R1C7. Naively, it could be 2, 4, or 8, but we can do better.
3b. Consider that the two other digits in the 11-cage are paired with the 69 pair in the 15-cage, and yet must sum to the same value. Since 6 and 9 are three apart, so too must the two other digits in the 11-cage. That eliminates 8, since 1 and 2 are too close together. In fact, of the five possible ways to make 11 out of the three distinct digits, one is all odd digits, two require a 6, and now we know 128 won't work, so the 11-cage is 245, and 4 is the leftover, since it's too close to both other digits. So that's the first digit: R1C7 is 4.
3c. We can fill out the rest of box 3 with pairs, since the uncaged zipper pair must be 38, leaving the remaining two cells as a 17 pair.

4. Let's look at box 2 now. The caged cells sum to 24, so row 1 sums to 21. Now it does so without including a 4, so they are either 579 or 678, but definitely include a 7. That resolves the 17 pair in box 3.

5. The zipper line that is entirely within box 2 has an odd number of cells, so R1C5 is the zipper value. It cannot be 5, since there are only two ways for pairs to sum to 5, and the two possibities for the 9-cage either provide only one of the needed digits, or consume three of them, leaving not enough for the 15-cage.
5b. It cannot be 9, because the 9-cage needs a zipper pair plus one more cell to make 9.

6. The 15-cage cannot include a 3, since that's used by the 9-cage. It cannot include a 7, since that's used by the top row. It cannot include both 4 and 5, since that would leave nothing for the 9-cage. It cannot include both 2 and 5, since that would leave nothing for R2C7. We're down to a few possibilities for the cage, all of which put either 8 or 9 into R2C5, with the other two digits adding up to either 6 or 7, eliminating 8 from R1C5.

7. Now where does 2 go in row 1? It can only go in box 1, and that eliminates 2 in the lower two-third of the 9-cage in box 1.
7a. Without a 2 below it, there is no longer a place for 6 in the 9-cage, given us three more digits: R1C3 is 2, R2C3 is 3, and R3C3 is 4.
7b. That resolves the order of the 15 cage, so R3C1 is 8 and R3C2 is 7.
7c. It also resolves the 38 pair in box 3, so R1C8 is 3 and R2C9 is 8.

8. The 8 in box 3 forces R2C5 to be 9, which eliminates 6 from the 15-cage, since that would reach 15 without using the third cell in the cage.
8b. That means the zipper value in R1C5 is 6, and it is flanked by a 78 pair on row 1.
8c. The just-placed 6 means the 3 on row 3 must be in R3C5.

9. Back in box 1, the four uncaged cells are 1569, and we need the two on the line to sum to 11, which resolve all four cells. R1C1 is 9, R1C2 is 5, R2C1 is 6, and R2C2 is 1.

10. That means the 15-cage in box 2 contains 24 and R2C7 is 5.
10b. That means the 9-cage in box 2 contains 15 and R3C7 is 2.

11. We can finish the zipper line in box 3 now: R3C8 is 9 and R3C9 is 6.

12. Time to move down the puzzle! Let's start with cages.
12a. The 19-cage in box 8 cannot have 6 or 9, and so must be 478. We'll come back to that one later, but it means the other three cells in column 5 are 125. We don't yet know which two of those are zipped together, so R5C4 could be 3, 6, or 7.
12b. The 11-cage in box 6 would normally have four possible pairs, although the placed digits in box 3 dictate the order of all four pairs. Additionally, R5C8 being 2 would require 1 in two cells of column 7, so the 29 pair is eliminated completely, meaning R5C7 could 3, 6, or 7.
12c. The 9-cage is box 4 works simililarly: although there are four possible pairs adding up to 9, the placed digits in box 1 dictate the order of all of them. Additionally, the 36 pair would put 7 in both R5C4 and R5C7, which won't work.
12d. Can the 9-cage actually be a 27 pair? Consider the effect that would have on box 5: Since R5C5 could not be 2, that would force 2 onto the zipper line centered at R5C4. The 2 would need to be paired with 1 to avoid putting 7 in R5C4, so 3 would go into R5C4 instead, and 5 would go into R5C5. Now there is no possible value for the 11-cage, since 3 and 7 aren't available, and 6 would require a second 5 in R5C8.

13. What's the value of R5C1? It's the sum of two cells in column 2, where there is already a 1. So at a minimum it is 5 (2+3), but 6, 8, and 9 are already in column 1, so it can only be 5 or 7. We've already seen that 5 would be the result of 2+3, and now 7 would need to be 3+4, since 1 and 5 already exist in column 2.
13b. Now consider the 9-cage again: If R5C1 is 5, then the 9-cage must be 81. If R5C1 is 7, that must come from 3+4, so again, the 9-cage must be 81. Either way, R5C2 is 8 and R5C3 is 1.

14. That eliminates 1 from R5C5, pushing it onto the zipper line that's fully in box 5. Now that line's value is either 3 or 6.
14b. Can the 11-cage in box 6 still be 65? That would force R5C4 to be 3, while also forcing R5C5 to be 2, denying the zipper line what it needs. So R5C7 is 7 and R5C8 is 4. That 4 on the center of a zipper line makes R4C7 and R6C7 a 13 pair.
14c. The just-placed 7 also puts 5 in R5C1, so R4C2 and R6C2 are a 23 pair.
14d. The 5 puts 2 in R5C5, making the cells immediately above and below it a 15 pair, and R5C4 therefore 6.
14e. There remain two empty cells in row 5, and R5C9 cannot be 3, so it is 9, while R5C6 is 3.

15. We have four empty cells in boxes 4, 5, and 6, so let's fill those in with candidates: 4679 in box 4, 4789 in box 5, and 2568 in box 6.
15b. We already have 6 and 9 in column 1, so we can split those quad cells into pairs: 47 and 69. Nice.
15c. We already have 6 and 8 in column 9, so we can also split those quad cells into pairs: 68 and 25.
15d. The zipper line in box 6 with a 9 on it is lopsided. The pair closest to the center of the zipper line can sum to either 11 or 14, but the other pair can't reach that high. It can be 8, 10, 11, or 13. The common sum is 11, so we can start to unwind rows 4 and 6 by placing 2 in R4C9.
15e. Now R4C2 is 3, R4C7 is 1, and R4C5 is 5. Also, R6C9 is 5, R6C5 is 1, R6C7 is 3, and R6C2 is 2.
15f. To balance out the zipper line in box 6, R4C8 is 6, R4C3 is 9, R6C3 is 6, and R6C8 is 8.

16. Let's resolve row 4 by considering the zipper line which has two digits on it already. The existing 2 and 3 are a single digit apart, and on opposite sides of the zipper line, so the other two digits must also be consecutive. That eliminates 4, and balancing the line puts 8 into R4C4 and 7 into R4C6.
16b. R4C1 is 4, and R6C1 is 7.
16c. We can also resolve row 1: R1C4 is 7 and R1C6 is 8.

17. Moving down again, it's time for the bottom third! We might as well start be filling in all the triples in boxes 7 and 9, where we definitely have triples. From left to right: 123, 469, 578, and skipping over to column 7: 689, 125, and 347.
17b. The zipper line extending in box 9 with a value of 8 can't have 4 on it, since that would require a second 4 in the box. It also can't have a 2 on it, since 6 isn't available for the other side. That means R9C8 is 2 and R9C9 is 4.
17c. Those are two digits apart, with the smaller digit closer to the center of the zipper line, so we can place 8 in R8C7 and 6 is R7C7 to balance out the line, and then 9 in R9C7 since it's right there.

18. Back in box 7, we've just knocked out two digits from a triple, forcing 6 into R9C2.
18b. The lack of negative numbers in Sudoku mean 7 and 8 are now too high for R8C3, which is 5.
18c. Balancing out that zipper line, R9C1 is 1. Speaking of 1, R8C8 is 1 since it's not 5.
18e. Balancing out that zipper line R8C9 is 7. That puts 3 in R7C9 and 5 in R7C8.

19. We've placed 7 and 8 in row 8, so R8C5 is now 4. That does a few things!
19b. It puts 3 into R9C4 and 1 into R7C4 to complete that zipper line.
19c. It also puts 6 into R8C6 and 9 into R7C6, since that's the only way to satisfy that zipper line now that its value is known to be 15.

20. Let's jump back up and use the just-placed 1 and 9 to resolve the top two-thirds of the grid: R6C6 is 4, R2C6 is 2, R2C4 is 4, R6C4 is 9.
20b. Now the other way: R3C4 is 5, R3C6 is 1, R9C6 is 5, and R8C4 is 2.

21. Row 8 has a couple of those ever-popular "naked singles": R8C1 is 3 and R8C2 is 9.
21b. R7C1 is 2 and R7C2 is 4.

22. One last use of a zipper line: R7C3 needs to pair with a 4 for a value of 12, so it's 8, R9C3 is 7, and I'll leave disambiguation of the final pair up to you.