Let $A, B, C, D$ be labeled in the clockwise manner. Let $a_n, b_n, c_n, d_n$ be respectively the number of ways to start $A, B, C, D$ and exit from $A$ in $n$ jumps. Then $b_n = d_n$. Since from $A$, the frog can return to $A$ in an even number of jumps, $a_n = 0$ when $n$ is even.

From $A$, in 1 jump, the frog can exit from 2 sides or jump to $B$ or $D$, Thus $a_1 = 2, a_3 = 4$. Therefore for $n \ge 3, a_n = 2b_{n-1}$.

From $B$, the frog can jump to $A$ or $C$. So $b_n = a_{n-1} + c_{n-1}$.

From $C$, the frog can jump to $B$ or $D$. So $c_n = 2b_{n-1}$.

Therefore
$$a_n = 4a_{n-2}.$$
Consequently, $a_{2m+1} = 4^{m-1}a_3 = 4^m$ for $m \ge 1$.