suppose u is in null(g). then fg(u) = f(g(u)) = f(0) = 0. so null(fg) contains null(g), so its (null(fg)'s) dimension has to be at least as big (as null(g)). (they could be the same, for example, g could be the 0-map).

on the other hand, u in null(fg) means g(u) is in null(f)∩g(U), which is perhaps smaller than null(f) (there might be v in V with f(v) = 0, but that aren't images g(u) for some u in U).