Nice puzzle, I certainly bumped my head along the way quite a few times. I hope you like the steps I took to solve it, as I enjoyed figuring it out.
5 cannot add with any other number to be 5, and any two connected 'w' tiles are adjacent, so lets start by filling in the empty 5 on r5c8, and the other 5s that follow in columns 7 and 9
An interesting thing to note is that a chain of 2 'w' connections in a single block/row/column will require that the middle of those 2 connections be less than 5. Proof by contradiction, if a square connected by two 'w's in the same block/row/column has a number > 5 (e.g. 8), then it only has one corresponding number to pair up with (e.g. 2). This would force two of those corresponding numbers to end up in the same block/row/column, which is illegal.
Now follow this logic with chains of 3 connections, we can then assert similarly that the two squares in the middle must both be less than 5. This helps us get some easy possible pairs in the 3rd block of the center row. The number 2 is ruled out, and likewise it's counterparts 3 and 8 are also ruled out from this 4-square chain.
This forces the chain to be built upon 14 in the center, with the outer set being a 69
Continuation of the w chains into the middle block, from r5c7, means that r5c6 must be a 14 - however, we already have a 1 in row 5, therefore, it must be a 4, which forces the rest of the sets of the chain to resolve.
The remainder of squares in the center left column give us an 8 and a 37 pair as we only have the options of 378 left, and we must make a w pair with one of those sets (only 37)
There is another 4 chain in the top right block. The 2 squares in c7 must be a 14 pair or a 23 pair, but there is a 4 in the column already, so it must be a 23 pair. Likewise, their paired tiles in c8 must be an 87 pair. This retroactively forces the 37 pair in r6 to resolve to 7 3, as 3 is no longer permitted in r6c7
Row 6 has only 4 remaining numbers, a pair of which must sum 5 or 10, and the options are 2456. Of those, the obvious answer is that the 46 must pair. Given that the 4 is tied up here at the bottom, the number paired with 1 on r5 is a 9.
Additionally a late follow-up to the logic in step 1. We know that the 5 in c1 must be in r4 or r6, as no other squares in that block permit a 5. Therefore, the bottom left square must have a 5 in c3, therefore the top left square must have a 5 in c2. The only unentangled spot in c2 is r2. Additionally, there must be a 5 in r1c6, just by nature of it being the only available square.
The only remaining numbers on row 5 also happen to tie into a web of entangled w-connections. These numbers are 378, of which 8 is the outlier. Following that chain of connections gets us the following options.
r4c3 is tied twice within the same row and block, therefore must be a low number. It also forces r4c4 and r5c3 to be the same value. This doesnt do much, but it does guarantee us that a 3 is in c3 at r4 or r5. It also forces a 2 into either r4c2 or r4c3. At this point in r4 we definitely have a 2 and a 3 in the entangled squares. If we focus in on r4c5, we can see that its not a 2, 3, or 7, and must be a 5. This forces the 25 pair in row 6 to resolve. This means that r4c1 must be a 7 or 8.
Lets resolve r4c1 - rules state that unentangled squares cannot sum to 5 or 10. If r4c1 was an 8, r4c2 would be forced to be 2, an illegal move as it is adjacent to an 8 unentangled. This forces r5c3 to be a 3, as it is the only remaining option, and forces the whole entangled set to resolve. Note, we still have a 38 unresolved in r3c1.
Lets detour back to the bottom-right corner real fast. With c7, 2, 3, and 4 are already consumed. This leaves only one option, a 19 pair. It gets more interesting when we then consider that there is a chain of 4 entangled squares in there as well. This chain must use a pair of low numbers to sum to 5 as the center two squares. The only remaining low numbers being 2, 3, and 4, it must consume 23. This resolves quickly as c7 only has 1 unaccounted for number - 8. This resolution of numbers forces the empty entanglement on c8 to be 46, conveniently forcing the 49 pair on r3c8 to be just 9. Before we stop here, we also must note that there are only 3 remaining numbers in c9: 1, 4, and 6. There's also one entanglement left, and it can be resolved by 14 or 46. 4 cannot go into one of the entangled tiles, so that leaves a 4 in r3c9, and a 16 pair across the rest.
Back at the top left, there's an unresolved chain of 3 entangled squares; however, we conveniently can rule out 2 and 3. So the chain must be either a 146 chain or a 419 chain. We know that the 4 must be at the top, for sure. Both r1c3 and r2c3 cannot be a 46, as that would break the fixed pair on r6 (at c3). Additionally, r1c2 cannot be a 9.
So, starting with r1c3 being 14, with the given restictions, it chains out as follows:
Doesn't take a keen eye to see the conflict on r3c4, resolving the 19 to be a 1, as r3 already has a 9. This forces the resolution to chain backwards. This also resolves the 46 pair in r6.
Notably, we can also now force the 9 to be r1c1, 7 to be in r3c3, and a 38 pair in r2c1, by sudoku. The 1 and 6 also resolve the 16 entanglement in the top right. Lastly, the 6 in the top center is forced to be paired with the 4, both by sudoku and by rules of entanglement eliminating the option of 1.
r1c4 is now critical, and will resolve all remaining entanglements in the top blocks. Note that the only remaining numbers for top-center are 2378, akin to the top-right's entanglements. Well, c4 already has 2, 3, and 7. r1c4 can only be 8, forcing it's pairing to be a 2, and resolving the chain on the right, which will resolve the 38 pair on the left afterwards. The final empty entanglement in the top center also resolves as 37.
Fantastic, we're almost done. Lets turn our attention to c1 and c2's remaining numbers and entanglements. C1 needs numbers 2, 4, and 6, while c2 need 1, 3, and 7. C1 can only support the entanglement of 4 and 6, 2 forced to r7c1. C2 can only entangle 3 and 7, 1 forced to r9c2. Additionally, r9c1 and r9c2 must be entangled as well - of 4 and 6, only 4 can entangle with 1. Plus, we have a 7 on r8 which forces the 37 pair to be oriented.
The previous position of 46 also fixes the 46 in the bottom right. The last entanglement left is in the bottom center, which should help us get everything we need. C6 requires numbers 2, 8, and 9 - of which 28 must be entangled. 9 must go in r9c6.
R9 still needs 5, 6, and 7. 7 is ruled out of c3 and c4, therefore it fills r9c5. 6 is ruled out from c3, so it fills r9c4, leaving 5 in r9c3. Sudoku then forces 5 into r8c4. r8 is still missing 1289, despite having some related entanglements.. However, r8c5 cannot be 9, 8, or 2, as c5 contains these. It must be 1, fixing the 19 pair on c7.
By noting that r8 requires a 28 only, and c3 already has a 2, this puzzle's remaineder is solved by natural sudoku rules.