Let us define $f(s, h)$ as the number of walks to get to the target vertex in $h$ hops, provided we're currently $s$ hops away (oriented, so $s$ can be negative), but instead of a polygon we have a line (this is analogous to $n = \infty$). Then, it is not hard to see that
$$
f(n, s, h) = \sum_{k=-\infty}^{\infty} f(s + kn, h).
$$
(Only finitely many of the summands will be nonzero, assuming $n > 0$.) This is just accounting for walks which go around the polygon a certain number of times. Now we will focus our attention towards finding an explicit formula for $f(s, h)$. We have the recurrence
$$
f(s, h) =
\begin{cases}
1 & h = 0, s = 0 \\
0 & h = 0, s \neq 0 \\
f(s+1, h-1) + f(s-1, h-1) & h > 0.
\end{cases}
$$
This recurrence has solution
$$
f(s, h) = \begin{cases}
\binom{h}{(h+s)/2} & h = s \bmod{2} \\
0 & h \neq s \bmod{2},
\end{cases}
$$
where $\binom{n}{r}$ is defined to be $0$ for $r < 0$ or $r > n$.
Combining this with the summation from before, we have a couple cases for $n$ even or odd and $h - s$ even or odd. In all of these cases, we want to compute a summation of the form
$$
\sum_{k=-\infty}^{\infty} \binom{h}{x+ky},
$$
where $x$ is either $\frac{h+s}{2}$ or $\frac{h+s+n}{2}$ and $y$ is either $\frac{n}{2}$ or $n$ depending on the parities mentioned above. We can compute this summation using any of a number of strategies, such as a root of unity filter; in particular, if we let $\omega$ be a primitive $y$-th root of unity, then the solution can be expressed in the form
$$
\sum_{j=0}^{y-1} \omega^{-jx} (1 + \omega^j)^h.
$$
This can't be reduced further easily, but it's closer to a closed form than nothing.
