If I tell you the moves for a game of chess and declare who wins, why can't it be checked in polynomial time if the winner does really win? This would make it an NP problem from my understanding.
First of all: The number of positions you can set up with 32 pieces on a 8x8 field is limited. We need to consider any pawn being converted to any other piece and include any such available position, too. Of course, among all these, there are some positions that cannot be reached following the rules of chess, but this does not matter. The important thing is: we have a limit. Lets name this limit simply MaxPositions.
Now for any given position, let's build up a tree as follows:
I'm now too tired to think of if we need one additional level of depth or not for the idea (proof?), but heck, just let's add it. The important thing is: the tree constructed like this is limited.
Next step: Of this tree, remove any sub-tree that is not reachable from the root via legal chess moves. Repeat this step for the remaining children, grand-children, ..., until there is no unreachable position left in the whole tree. The number of steps must be limited, as the tree is limited.
Now do a breadth-first search and make any node a leaf if it has been found previously. It must be marked as such(!; draw candidate?). Same for any mate position.
How to find out if there is a forced mate? In any sub tree, if it is your turn, there must be at least one child leading to a forced mate. If it is the opponents move, there must be a grand child for every child that leads to a mate. This applies recursively, of course. However, as the tree is limited, this whole algorithm is limited.
[sensored], this whole algorithm is limited! There is some constant limiting the whole stuff. So: although the limit is incredibly high (and far beyond what up-to-date hardware can handle), it is a limit (please do not ask me to calculate it...). So: our problem actually is O(1)!!!
The same for checkers, go, ...
This applies for the forced mate, so far. What is the best move? First, check if we can find a forced mate. If so, fine, we found the best move. If there are several, select the one with the least moves necessary (still there might be more than one...).
If there is no such forced mate, then we need to measure by some means the 'best' one. Possibly count the number of available successions to mate. Other propositions for measurement? As long as operating on this tree from top to down, we still remain limited. So again, we are O(1).
Now what did we miss? Have a look at the link in your comment again. They are talking about an NxN checkers! The author is varying size of the field!
So have a look back at how we constructed the tree. I think it is obvious that the tree grows exponentially with the size of the field (try to prove it yourself...).
I know very well that this answer is not a prove for that the problem is EXP(TIME). Actually, I admit, it is not really an answer at all. But I think what I illustrated still gives quite a good image/impression of the complexity of the problem. And as long as no one provides a better answer, I dare to claim that this is better than nothing at all...
Addendum, considering your comment:
Let me allow to refer to wikipedia. Actually, it should be suffient to transform the other problem in exponential time, not polynomial as in the link, as applying the transformation + solving the resulting problem still remains exponential. But I'm not sure about the exact definition...
It is sufficient to show this for a problem of which you know already it is EXP complete (transforming any other problem to this one and then to the chess problem again remains exponential, if both transformations are exponential).
Apparently, J.M. Robson found a way to do this for NxN checkers. It must be possible for generalized chess, too, probably simply modifying Robsons algorithm. I do not think it is possible for classical 8x8 chess, though...
O(1) applies for classical chess only, not for generalized chess. But it is the latter one for which we assume not being in NP! Actually, in my answer up to this addendum, there is one prove lacking: The size of the limited tree (if N is fix) does not grow faster than exponentially with growing N (so the answer actually is incomplete!).
And to prove that generalized chess is not in NP, we have to prove that there is no polynomial algorithm to solve the problem on a non-deterministic turing machine. This I leave open again, and my answer remains even less complete...