assuming I have two matrices and there SVD's
$$ A = U_A \Sigma_A V_A^T \\ S = U_S \Sigma_S V_S^T $$
does that necessarily mean that $$ \begin{bmatrix} A \\ S \end{bmatrix} = \begin{bmatrix} U_A & 0 \\ 0 & U_S \end{bmatrix} \cdot \begin{bmatrix} \Sigma_A & 0 \\ 0 & \Sigma_S \end{bmatrix} \cdot \begin{bmatrix} V_A^T \\ V_S^T \end{bmatrix} $$
is one of the possible SVD of $\begin{bmatrix} A \\ S \end{bmatrix}$ ?
No, because your matrix of right singular vectors is $[V_A, V_S]$ is not unitary. If $A$ and $S$ have $N$ columns, then $[V_A, V_S]$ is going to be $N \times 2N$ which cannot be a unitary matrix. To find a suitable $V$, you'd need the eigenvectors of $A^T A + S^T S$ or so.
That said, $\begin{bmatrix}U_A & 0 \\ 0 & U_S\end{bmatrix}$ does qualify for a matrix of left singular vectors, as it is unitary and spans the right space.
The singular values are influenced by both $U$ and $V$ so they will change as well.