The following screenshot is taken from the Android app Kakuro++ (level 9 riddle 12)
And I verified the current status within the app. I managed to solve the puzzle by guessing a 50/50-number and eventually having to backtrack to pick the other possibility. However, I'm not a big fan of that approach. I'm looking for a trick that can preferably be applied also to other puzzles of similar difficulty.
In the bottom right corner, in the 6-clue pointing right, you know:
The leftmost square cannot be a 1, since this would force both the bottom squares in the 3 and 4 clues to its right to be a 1.
This forces:
The square diagonally above the 6-clue to be a 1. Then, the 12-clue in this column must be 1245, since there is no longer a place for the 6 to go. This lets you fill out the column like so:
The bottom right corner falls to easy deductions:
Next:
Look at the 14-clue down in the middle of the puzzle. We now have a 3&5 in the column, so there cannot be a 6 in the top cell. This forces the crossing 12-clue to be 1245, and thus the cell directly under the 14-clue is 4. With some easy consequential fill-ins:
Finishing up:
In the 15-clue down in the fifth column, the remaining two clues need to sum to 10, which force them to be 8 and 2. The rest is just fill-ins.
You can start in the area near the orange cursor:
The vertical 3 and 4 to the right of the cursor both use a 1. Therefore there is a 1 on each row there, and the square above the orange cursor cannot be a 1.
The vertical 12 must use a 1 somewhere (2+3+4+5>12) and there is only one possibility left for that.
(Dupe, but with pic)
You have an X-wing of ones at the bottom right, so you can place the 1 in the 12 clue:
