Singular value decomposition of product of matrices

Actually there is an answer, but somewhat complex. Denote the SVD decomposition of $A$ by $U^*S^*V'$, denote $\operatorname{qr}(BU) = QR$, and denote SVD decomposition of $RS$ by $U_1^*S_1^*V_1'$ then the SVD decomposition of $BA$ is $(QU_1)S_1(VV_1)'$.

Proof: $BA= BUSV'=QRSV'=(QU_1)S_1(V_1'V') = (QU_1)S_1(VV_1)'$. $QU_1$ is unitary and $VV_1$ is unitary and $S_1$ is diagonal.

Interpretation: the singular values of $BA$ are the singular values of $RS$. So the original singular values are multiplied by the upper triangular $R$ of the qr decomposition of $BU$, i.e., $B$ after unitary transformation by right bases $U$ of $A$.

There really isn't a simple relationship between the SVD of a product and the SVD of the individual factors.

However, there are methods for forming the SVD of a product of two or more matrices, without forming the matrix product itself (which can be a source of inaccuracy); see for instance this paper by Golub, Solna, and van Dooren.