Time complexity of BFS with min heap
Yes.  Both bounds are valid.  There's not much difference between the two bounds.

There are certainly at most $mn$ items in the queue at any time, as there are only $mn$ possible items.  Consequently, the running time is at most $O(mn \log(mn))$.

By analyzing how BFS visits nodes and the size of the "fringe", one can show that the size of the fringe is most $m+n$.  In particular, the fringe is a path that goes from somewhere on the top of the matrix to somewhere on the left of the matrix, by going only down, left, or diagonally down-and-left (but never right or up).  Thus the fringe can include at most $m+n$ cells.  This means that at every stage, the size of the heap is at most $m+n$, so $O(mn\log(m+n))$ is also a valid upper bound on the running time.

These two bounds are pretty close, so either is a reasonable upper bound on the running time.

Caveat: there are multiple heap data structures, with different running times for the heap operations.  This answer assumes you are using a heap data structure where inserting into a heap containing $k$ items can be done in $O(\log k)$ time.