This is a great hybrid puzzle, which required a fair bit of thought to solve.
[Note that this answer fails to take advantage of the clue-adjacency rule, which it turns out is not required.]
First, the final grid:
And the path to get there. (Apologies if some of this is hard to follow. It's quite difficult to describe in words some of the patterns/restrictions that emerge when you analyse the grid and the combined ruleset) :
I assumed (not stated here, but standard for a numberlink puzzle) that there must be one numberlink line per distinct digit, so exactly two of each digit (excluding 0) are numberlink clues, and the rest are slitherlink.
There are only two 6's so they must be numberlink clues. They are both on an edge of the grid, so the line between them will bisect the grid. (The slitherlink loop cannot squeeze past on the outside edge of the grid because of the 0 adjacent to one 6.)
So the slitherlink line must all be on one side of the 6-line.
And the other numberlink clues must always fall (pairwise) on the same side of the 6-line.
There are four 5's, so two must be slitherlink clues and must be on the same side of the 6-line. A numberlink line cannot pass through a slitherlink 5-cell.
There are also four 4's, so two numberlink and two slitherlink clues; and each pair must be on the same side of the 6-line so we have a 0-4 or 2-2 split.
A numberlink line can pass through a slitherlink 4-cell, but either enters and exits through adjacent edges, or passes between two segments of the slitherlink line. But we have three 4's adjacent near the bottom, and the 6-line cannot pass through one them without bisecting the slitherlink loop. So the 6-line must pass above them, and must therefore pass above (possibly through) the other central 4-cell.
And we now know that the 6-line is entirely above the slitherlink loop (because all the 4's are below the 6-line).
The 6-line cannot get above the two 5's on the edge of the grid, so those must be numberlink clues. And the other 5's must be slitherlink clues, below the 6-line.
We can finally start filling in some lines (colouring known numberlink and slitherlink clues green and blue respectively).
Now, there are three 2's, so two are numberlink clues, and the other is a slitherlink clue that must be below the 6-line. The 6-line cannot pass completely above any 2-cell without blocking the 5-line, so the 6-line must pass through the slitherlink 2-cell. But that is only possible with the lowest 2:
And that gets us to here:
Continuing with basic deductions gets us to:
The central 4 cannot be a slitherlink clue, as it would close the loop and violate the slitherlink rules for the adjacent 5, so it must be a numberlink clue.
Continuing with basic deductions:
The rightmost 1 cannot be a slitherlink clue, as the loop would cover two edges, or block the 6-line. So it must be a numberlink clue.
And more basic deductions get us to:
Note the slitherlink edge marked red. This is the only path for the slitherlink loop to escape the lower part of the grid, so must be part of the loop.
And the loop has to cover at least two edges of the 1 above, so that 1 must be the other end of the numberlink 1-line.
From there, basic deductions get us almost all the way:
From there, I ran out of logic...
Assuming that the upper remaining 4 was the other end of the 4-line leads to a contradiction in the slitherlink path around the remaining 4's.
So that 4 must be a slitherlink clue, and the loop must go round it clockwise to avoid completing the loop prematurely.
Then the next 4 down cannot be a slitherlink clue, as that would cut off the final 4 from the 4-line.
So the lowest 4 is the final slitherlink clue, and basic deductions give us the complete grid.