what is the relationship between the SVD of $[A;B]$ and that of $A$ and $B$? In my question, $A\in R^{m\times r}$, $B\in R^{n\times r}$, and $[A;B]\in R^{(m+n)\times r}$ results from stacking $A$ over $B$. Given $m>r$ and $n>r$, we do the SVD on $A$ and $B$ and have that $A=U_aS_aV_a^T$ and $B=U_bS_bV_b^T$, where $U_a\in R^{m\times r}$, $S_a\in R^{r\times r}$, $U_b\in R^{n\times r}$, and $S_b\in R^{r\times r}$. 
Can we use the above results to get the SVD of $[A;B]$?

Not really, even take the simplest case of $1 \times 1$ matrices. 


*

*Let $A= \begin{bmatrix}1 \end{bmatrix}$ and let $B=\begin{bmatrix} 2 \end{bmatrix}$. 

*Singular value of $A$: $\sigma_a = 1$

*Singular value of $B$: $\sigma_b = 2$

*SVD of $[A ; B] = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$: $$U S V^T = \begin{bmatrix}
 \frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\
 \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\
\end{bmatrix} 
\begin{bmatrix} \sqrt{5} \\ 0 \end{bmatrix} \begin{bmatrix} 1 \end{bmatrix}$$


There is a slight relationship on the singular value, you'll note that $\sqrt{5} = \sqrt{ \sigma_a^2 + \sigma_b^2}$. However, this is really only an artifact of $\mathbf{v} = \begin{bmatrix} 1 \end{bmatrix}$ being an eigenvector for $A$, $B$, and $[A; B]$; the interplay is generally much more complicated. Note that since $U$ and $V$ are rotation matrices, the singular value decomposition doesn't play particularly well with most block structures.
For a $2 \times 2$ matrix, consider $A= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ (singular values of $1$) and $B= \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ (singular values of $2$). The SVD of $[A;B]$ is 
 $$USV^T = \begin{bmatrix} 
 0 & \frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} \\
 \frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} & 0 \\
 0 & \frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
 \frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} & 0 \\
\end{bmatrix}
\begin{bmatrix} 
 \sqrt{5} & 0 \\
 0 & \sqrt{5} \\
 0 & 0 \\
 0 & 0 \end{bmatrix}
\begin{bmatrix}
0 & 1 \\ 1 & 0 
\end{bmatrix}. $$
It only gets worse if you make $A$ and $B$ "messier." Again, this is an example where $A$ and $B$ share all of their eigenvectors. 
