If $A$ is a rectangular matrix of dimensions $m\times n$, then $S_L=AA^T$ and $S_R=A^TA$ are square symmetric matrices. Hence, using the eigendecompostion we can write $$ S_L=AA^T=U\Lambda_{S_L} U^T $$ and $$ S_R=A^TA=V\Lambda_{S_R} V^T $$ Where, $U$ is the matrix of eigenvectors of $S_L=AA^T$ and $V$ is the matrix of eigenvectors of $S_R=A^TA$ and $\Lambda_{S_L}$ is the diagonal matrix of eigenvalues of $AA^T$ and $\Lambda_{S_R}$ is the diagonal matrix of eigenvalues of $A^TA$.
My question is how can we go from this result to singular value decomposition which says $$ A=U\sqrt{\Lambda}V^T $$ ?
The problem here is that $U$ and $V$ are chosen independently from two different, but related, spectral decompositions. To ensure that they are compatible in the sense that $A = U\Sigma V^T$, one has to choose one basis from the spectral decomposition, then construct the other from $A$.
For example, let $A^TA = V\Lambda V^T$, then notice that $AV = U\Lambda^{1/2} $, so $U$ is uniquely defined as $U = AV\Lambda^{-1/2}$. This argument works the same way when $A$ is not of full rank by using the pseudo-inverse of $\Lambda^{1/2}$, which can be defined without the SVD for diagonal matrices.