Complexity of "Subtree of Another Tree"
The time complexity of isSame is not linear in the height of the smallest tree. Think for example of two copies of a complete binary tree (where all nodes have the same value).

That said, the time complexity of isSame is always $O(m)$, where $m$ is the number of vertices in the smallest of the two input subtrees.

Let me assume that the number of vertices $m$ of $t$ is at most the the number of vertices $n$ of $s$ (as otherwise the answer is trivial).

Your algorithm requires time $\Theta(n m) = O(n^2)$ in the worst case.
The upper bound is obtained by simply multiplying the number of vertices of $s$ with the $O(m)$ upper bound to the complexity of isSame.
The lower bound is obtained, e.g., by letting $s$ be a path on $n$ nodes rooted in one of its endpoints and $t$ be a path on $m$ nodes rooted in one of its endpoints (as before, all vertices have the same value). The first $\lfloor n/2 \rfloor$ invocations of isSame require $\Omega(m)$ time each, for an overall time complexity of $\Omega(n m)$.
Notice that this is  $\Omega(n^2)$ for $m=\Omega(n)$.