If $\mathbf \Sigma$ is a nonsquare diagonal matrix,and the elements in there are all real positive value.
Now $\mathbf V$ is a orthogonal square matrix,which means $\mathbf V \mathbf V^H=\mathbf V^H \mathbf V=\mathbf I$.
Now there is a formula
$\mathbf U \mathbf \Sigma \mathbf V^H(\mathbf V \mathbf \Sigma ^2\mathbf V^H)^{-1}\mathbf V \mathbf \Sigma$,is my calculation right?
$\mathbf U \mathbf \Sigma \mathbf V^H(\mathbf V \mathbf \Sigma ^2\mathbf V^H)^{-1}\mathbf V \mathbf \Sigma=\mathbf U \mathbf \Sigma \mathbf V^H(\mathbf V \mathbf V^H \mathbf \Sigma ^2)^{-1}\mathbf V \mathbf \Sigma=\mathbf U \mathbf \Sigma \mathbf V^H\frac{\mathbf I}{diag(\mathbf \Sigma ^2)}\mathbf V \mathbf \Sigma=\mathbf U \mathbf Idiag(\mathbf \Sigma ^2) \frac{\mathbf I}{diag(\mathbf \Sigma ^2)} \mathbf V^H\mathbf V =\mathbf U ?$
A lot of the question doesn't make sense, but to answer the question in the prompt: NO. If you have $VD^2V^T$ this will not in general equal $VV^TD^2$. As you said, $V$ is orthogonal hence satisfying $VV^T = I$. A counterexample to $VD^2V^T = VV^TD^2 = D^2$:
$$ \left[ \begin{array}{cc} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \\ \end{array} \right ]\left [ \begin{array}{cc} 1 & 0 \\ 0 & 2 \\ \end{array} \right ]\left[ \begin{array}{cc} 1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2} \\ \end{array} \right ] \;\; =\;\; \left [\begin{array}{cc} 3/2 &1/2 \\ 1/2 &3/2 \\ \end{array} \right ]. $$