Why is the set of NFA that accept all words in co-NPSPACE?
Your question is really about the following statement: If an NFA with $n$ states does not accept every string, then it rejects some string of length at most $2^n$.  (The rest of the proof comes from the fact that REACH is in NL.)

To see why this is so, first let's look at a DFA with $k$ states that does not accept every string.  The transition function of the DFA corresponds to a directed graph $G$ on $k$ vertices. DFA computations bijectively correspond to walks in $G$ starting from the vertex $s$ representing the start state.

Since the DFA does not accept every string, there is a vertex representing a non-accepting state $j$ reachable by a walk from $s$.  In particular, there is a simple path $P$ from $s$ to $j$, which can have length at most $k$ (e.g., DFS will produce one).  $P$ corresponds to a string of length at most $k$ that is not accepted.

The NFA question reduces to the DFA one.  Using the powerset construction, convert the NFA with $n$ states to a DFA with $k = 2^n$ states and then use the above argument.