Sum-body Told Me...
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For solving: https://sudokupad.app/ixwox25caa

1. A three-cell segment on a Region Sum Line limits a single-cell segment on the same line to 6, 7, 8, or 9. There are three such instances in this puzzle. R3C6 is in a box with a given 6, so it is 789. R6C7 is in a box with a given 9, so it is 678. R7C4 shares a column with a given 6, and a box with a given 8, but we can do better; the three-cell run in box 7 cannot contain a 4, which is required for three cells to sum 7, so R7C4 is 9.

2. Now there is only cell is box 9 that can be 9. Within that box, row 7 and column 9 are ruled out by placed and given 9s, and three of the four remaining cells are part of multi-cell segments on lines with single-cell segments, and so cannot be the maximum digit. R8C7 is 9. After placing that, the same logic applies to box 3, where R1C8 is 9.

3. The candidates for R6C4 cannot include 1 or 2, since there's a 2-cell segment involved, and there's a given 6 and a placed 9 in the column, so the candidates are 34578. With that in mind, where can 8 be in box 7? Not on the 2-cell segment with a maximum sum of 8, and not on the 3-cell segment with a maximum sum of 9, and not on row 9 where there is a given 8. R8C1 is 8.

4. R7C7 has the same sum as R6C8+R6C9. Those two cells share a box and a row with 1 and 2, so their minimum values are 3 and 4, and their maximum values are 3 and 5 to avoid pushing R7C7 too high. Either way, one of them must be a 3, and R7C7 is either 7 or 8.

5. Let's focus on R6C4.
   5a. We just identified a "pointing pair" in which the 3 in row 6 is either in R6C8 or R6C9, so R6C4 is updated to remove 3 as a candidate, and now must be from 4578.
   5b. A 3-cell segment which sums to 9 has three options, but with the given 4 in the box, it has only two, both of which contains a 1 (126 or 135). That eliminates 4 from R6C4, which now must be from 578.
   5c. But wait! We just said that the 3-cell segment in box 7 is either 126 or 135. Either option eliminates both 14 and 23 from the 2-cell segment in box 7, and those are the only pairs which sum to 5, so R6C4 must now be from 78.
   5d. Can it be 7? The 2-cell segment would need to be 16, 25, or 34. The given 4 and the need for a 1 on the 3-cell segment eliminate two of those. If it were 25, that would again leave no value for the 3-cell segment, which must 126 or 135. So R6C4 cannot be 7, and is 8.
   5e. Continuing in box 7, there are three options for the 2-cell segment which now sums to 8. The 1 on the 3-cell segment means it cannot be 17. It could be either 26 or 35, but the 7 is now forced into row 9, where the given 7 in box 4 means R9C3 is 7, and that forces R9C1 to be 9.

6. Let's mark our remaining single-cell RSL segments, and catch up on pencil marks from prevous steps.
   6a. R6C7 is now 6 or 7. That means the 3-cell segment on that line in box 9 is either 123 or 124, and the given 1 and 4 further reduce the candidates in R9C7 and R9C8. The 7 we just placed puts 7 in either R2C3 or R3C2. R3C3 is then from 589, with the given 3 in the box eliminating both 3 and 4. So far, we don't know much about R4C6; it can be anything but 1 or 2.
   6b. The 8s in the bottom rows force 8 in row 7 into box 9, but an 8 in R7C9 would require R7C7 to be at least 9, which it cannot be. So 8 is in either R7C7 or R7C8. There are multiple places 8 could be in box 6, but none of them are in column 9. Specifically, if R4C9 were 8, keeping the 2-cell segment sum to the maximum would require R5C8 to be 1, which it cannot. So the 8 in column 9 must be in neither box 9 nor box 6, nor in R3C9 where it would push R3C6 too high. 8 is in either R1C9 or R2C9.
   6c. While we're pencil-marking candidates, there are four pairs that add to 9. But R6C1+R5C2 are in a box with a given 7, eliminating 27, and R5C2 is in a column with a given 3, while R6C1 is in a row with a "pointing pair" for 3, eliminating 36. We're down to two possible pairs, and existing given and placed digits define the order. R5C2 must be from 58, while R6C1 must be from 14.
   6d. With that in mind, let's see how we can find 8 in column 7. Since we worked out earlier that the 8 in box 3 must be in column 9, and we have a given and a placed digit in column 7 already, with two other cells on rows which already contain 8s, there are only two options: R5C7 or R7C7. To figure out which it is, consider: Where is 5 in row 6? It's either part of the 2-cell segment in box 6, which would make R7C7 to be 8, or it's one of R6C2 or R6C3 in box 4, either of which would force R5C2 to be 8. Either way, R5C7 is not 8, so R7C7 is 8, and R6C8+R6C9 are a 35 pair.

7. Since we worked out in step 6a that the 1 and 2 in box 9 are on the 3-cell segment, R8C8+R7C9 are also a 35 pair. That collapses the 3-cell segment by removing the 3 from it, so R9C7 is 2, R9C8 is 1, R8C9 is 4, and R6C7 is 7.

8. Only two digits are missing in box 9 (counting the 35 pair as not missing), and the placed 7 in row 9 means that R7C8 is 7 and R9C9 is 6. The two remaining cells in row 9 must be a 35 pair, and now the only option for the 2-cell line segment in column 6 is 16. The remaining three cells are from 247, and the 4 in box 9 means R7C5 is 4. Now the line on which we just placed 4 has a sum of either 6 or 11.

9. Both box 6 and column 9 have all odd digits placed, although the ordering of 3 and 5 is not yet resolved. The remaining cells in each are even: 2468. R4C9 shares a column with 4 and 6, and the "pointing pair" 8 in box 3, so it is 2. Now the line on which we just placed 2 initially seems like it can be 6, 8, or 10. But the same line contains a single-cell segment with a minimum value of 7, so R3C6 is 8, R5C8 is 6, R5C7 is 4, and R4C8 is 8.

10. A 3-cell segment summing to 8 must contain a 1, So R3C9 is 1, and since R2C8 is either 2 or 4, then R2C7 is either 3 or 5. In addition, R1C9 and R2C9 are now a 78 pair. Whether you observe that R1C7 shares a row or column with every digit except 6, or that boxes 2, 6, and 9 together force 6 into two cells in box 3, one of which we just determined must be 3 or 5, either way: R1C7 is 6, and R3C7 is either 3 or 5.

11. Just moments ago, R4C6 could be any one of six different digits. Now that line's sum is made up of either 25 or 34, whichever is not combined with 1 on the 3-cell line in box 3, so R4C6 is 7.

12. Since we placed 8 in step 9, R3C3 is now from 59. Except there is already both 1 and 2 in row 4, so there is no possible 2-cell segment that adds to 5, so R3C3 is now 9.

13. Straightforward elimination helps at this point. Rows 4 and 6 both include 28 in boxes 5 and 6, so in box 4 they must appear in row 5, and we worked out earlier that R5C2 can't be 2, so R5C3 is 2 and R5C2 is 8. We need a 9 in column 2, and it can only be in R6C2. That forces the 9 in box 5 into R4C5, which forces 9 in box into R2C6, and that's all the 9s in the puzzle. Sticking with standard Sudoku rules, ignoring lines for now, 6 in boxes 2 and 6 mean R6C5 is 6. The 4 in box 6 pushes 4 into R4C4, and that in turn pushes 4 into R1C6.

14. Back to Region Sum Line logic: Since the 48 pair in box 6 mean that line's sum is 12, R5C6 and R3C5 are both 3. Since the 29 pair in box 4 mean that line's sum is 11, R5C4 is 5 and R8C4 is 7. The 9 in R7C4 defines that line's sum, so R1C6 is 1.

15. We could have done this back at step 7, once we worked out that R7C9 was either 3 or 5. That means the 2-cell segment in box 7, which sums to 8, cannot be 35, and so is 26. This now resolves the order of the 16 pair in column 6: R7C7 is 1, and R8C7 is 6. We can finish box 8: R9C6 is 5, R9C4 is 3, and R8C5 is 2. Some other leftovers: R5C5 is 1, R6C3 is 4, R2C5 is 7, and then R2C9 is 8 and R1C9 is 7. Also, R2C2 is 5, which displaces a penciled-in 7, so R3C2 is 7, while R3C1 is 4 by line logic.

16. That resolves box 3: R2C7 is 3, R2C8 is 4, R3C7 is 5, and R3C8 is 2.

17. There are four 8s eliminating all but one cell in box 1, so R1C3 is 8. Both R3C2 and R2C5 are 7, and both are part of 2-cell segments on the same line. Whether the sum of the line is 8 or 9, both R1C4 and R2C3 must be the same, and must be from 12. In row 2, then, R2C1 is 6.

18. R1C1 could be 1 or 2, but the 1 in box 4 means R1C1 is 2. That resolves the R1C4 and R2C3 digit as 1, and means R2C4 is 2.

19. In row 4, the 2-digit segment which sums to 9 cannot be 18, 27, or 45 because of existing digits, so R4C1 is 3, R4C2 is 6, and R4C3 is 5. The 6 we just placed resolves the 26 pair on row 7, so R7C2 is 2 and R7C3 is 6. There is only one remaining cell in column 2, so R8C2 is 1.

20. I left this for last just for fun. Resolving the remaining six 35 pairs is left as an exercise for the solver.