But if you were to ask the decision problem "Given an nxn board with a given position, can white win?" and apply it to generalized chess, would this variation be in NP?
Probably. In fact, I suspect it would be in logspace or perhaps even a smaller complexity class. Determining that white can win involves barely more than checking that the position isn't a stalemate, white isn't already checkmated and white has sufficient material to checkmate (from memory, a queen; a rook; two bishops; three knights; a bishop and a knight; or a pawn, via promotion). That doesn't quite work, since it could be that black's only legal moves unavoidably force checkmate or force the loss of sufficient mating material for white. But it feels pretty close to the truth.
Because the verifiability of this proof would only require polynomial time (just play out the winning game).
That's not true because there's no guarantee that the number of moves required is bounded by some polynomial in the size of the input.
For this decision problem, would it fall in NP, because a problem solvable by a DTM in EXPTIME and verifiable in polynomial time would be NP-complete, yes?
No, that's not the definition of NP-completeness. While it's a reasonable intuition that a problem being NP-complete means that we don't know any algorithms that are better than exponential, NP does not mean "it takes exponential time on a deterministic TM." Being NP-complete means that the problem is in NP and every other problem in NP reduces to it by, say, a polynomial-time many-one reduction.