There is no need to guess, also proves that solution is unique. L stands for left sudoku, R for the right one. (X,Y) means the X is in L, Y in R, the rest should be self-evident. Formatting of sudokus is crappy but I believe it should be clear enough.
First solve 10s and 11s. Note that 11 is 6+5, this prevents 10s in same rows (= all of them) from being 5+5. Additionally, focus on 10 in the second bottom row - that one has to be 4 in L because 5 and 6 are taken by 11s. This quickly gives solution for all 10 and 11 for both sudokus.
Then
Fill in 6 in the R, using ordinary sudoku rules. This is exactly on the position of left column top 2 of the combined sudoku, this also gives 1,1 for the other 2s. At this point we have the following image:
L/R, horizontal lines are not drawn.
x5x|6xx . . . x6|x4|xx
xxx|xxx . . . xx|xx|xx
x1x|xxx . . . x1|xx|xx
2xx|xxx . . . 6x|xx|xx
xxx|4x6 . . . xx|x6|x5
1x6|x5x . . . 1x|4x|6x
Then
Top left 5 can't be 1,2 or 5 because of L, it cannot be 1 in R, so it only has (3,2) option left. This enables filling in 5 in bottom left rectangle of the L, using ordinary sudoku rules, followed by 4 in same rectangle, 6 in top-left one, then 6 in mid left, 3 in mid left, 2 and 3 in bottom left, 4 in top left.
Solution of L/R at this point is
45x|6xx . . . x6|x4|xx
36x|xxx . . . xx|xx|xx
61x|xxx . . . x1|xx|xx
23x|xxx . . . 6x|xx|xx
523|4x6 . . . xx|x6|x5
146|x5x . . . 1x|4x|6x
Now notice that
2 is in position of 5 of combined sudoku, giving 3 for the R. The other 5 in the 2nd row cannot be 3,4 in L and 2,4 in R, so it has to be (2,3). Fill in the remaining 1 in top-left of L. The 2nd row 4 has to be 4 of the L, because 2 and 3 are blocked in both sudokus, preventing 1+3 or 2+2 combination. Then 9 has to be (6,3) because all other numbers from the L are taken.
At this point we have
451|6xx . . . x6|x4|xx
362|x4x . . . 2x|3x|xx
61x|xxx . . . x1|xx|xx
23x|x6x . . . 6x|xx|3x
523|4x6 . . . x3|x6|x5
146|x5x . . . 1x|4x|6x
Note
Top right 4 of combined sudoku has to be composite, and cannot be (1,3) because of L. This means 3 in top right of R has only one spot left. Fill in the other 3s. Then finish left column of R, then 4, bottom right 1 and 2, 2nd left column and remaining numbers.
This gives
451|6xx . . . 36|54|21
362|x4x . . . 24|31|56
61x|xxx . . . 51|62|43
23x|x6x . . . 62|15|34
523|4x6 . . . 43|26|15
146|x5x . . . 15|43|62
Only one part left now:
Top right 4 is composite and now we know it is 3 in L. This gives all other 3, and 2 in the top row with 1 in the bottom right part, then 2 in the bottom right part, 2 in mid right part. Now we know that the last number 5 of combined sudoku has to refer to the left one, because the remaining option (2,3) is not possible. This now solves the remainder of the sudoku (start with 5, then 4, then 1).
Solution is, as Earlien already found:
451|623 . . . 36|54|21
362|541 . . . 24|31|56
614|235 . . . 51|62|43
235|164 . . . 62|15|34
523|416 . . . 43|26|15
146|352 . . . 15|43|62