If $V$ is $f$, then it is finite (at least in one-dimensional case).

Who is $V(\cdot)$?

By the way, maybe the following strategy may be of use: since you can assume that all elements of the initial matrix are 0 or 1, you can brute-force this $m(n)$ on a computer at least for n=1,2,3,4,5 (maybe 6 too - there are $2^{n^2}$ matrices). And then look up that sequence on oeis.org

Just to be sure: you mean "meet for the first time at time $t$", or "be at the same location at time $t$"?

I mean, let us suppose that your vertices are rather labeled $\{0,\ldots,n-1\}$, and you start in $0$. Now, instead of considering the random walk with reflection in $0$ (and waiting until it comes to $n-1$), you may as well consider the simple random walk on $\{-(n-1),\ldots,n-1\}$ (in fact, the reflected random walk is just its absolute value). So, the hitting time of $n-1$ for the reflected random walk is the same as the hitting time of $\{-(n-1),n-1\}$ for the "normal" random walk on $\{-(n-1),\ldots,n-1\}$.

what's the initial position of the walk?

basically, yes. Seems we typed it roughly at the same time...