- Let $A,B$ be hermitian matrices. Since we consider proper elements of the matrices, we may assume that $B$ is diagonal; assume that we know $spectrum(A)=(\alpha_i),spectrum(B)=(\beta_i)\subset \mathbb{R}^n$ (in non-increasing order). What can we say about $spectrum(A+B)=(\gamma_i)\subset \mathbb{R}^n$ (in non-increasing order)?

The answer is given by the Horn's conjecture (cf. http://math.ucr.edu/~jdolan/schubert2.pdf ) whose proof is dated 1998-1999. The NS condition contains only one equality, the obvious one, $\sum_i\gamma_i=\sum_i\alpha_i+\sum_i\beta_i$ and many linear inequalities linking some $\alpha_i,\beta_i,\gamma_i$. Finally, for generic hermitian matrices $A,B$, the $(\gamma_i)$ satisfy only one equality OR the $(\gamma_i)$ goes throught a real algebraic subset of dimension $n-1$.

- Although $A,B$ are linked by $B=r.diag(A)$, it is of little importance because the spectra of $A,B$ are linked by only one equality, the obvious one. Indeed one has

Proposition (Schur-Horn): for every $p< n, \sum_{i=1}^p \beta_i\leq\sum_{i=1}^p\alpha_i$ and $\sum_{i=1}^n \beta_i=\sum_{i=1}^n\alpha_i$ IFF there is a hermitian matrix $C$ s.t. $diag(C)=(\beta_i)$ and $spectrum(C)=(\alpha_i)$.

- Application to your question. We consider the "thin SVD" $A=USV^T$ where $UU^T=I_n,V^TV=I_d$ and $S\in M_d$ is diagonal. Then $AA^T=US^2U^T$ (your formula is false), that is the standard orthogonal diagonalization of $AA^T$; in particular, I do not see the point to compute the SVD of $A$. The required formula is in the form $AA^T+r.diag(AA^T)=W\Sigma W^T$. We saw that there is no real link between eigenvalues of $S^2$ and $\Sigma$ and, consequently, between their eigenvectors $U,W$.

Conclusion. The answer to your question is NO and do not dream, there is no hope.