Singular value decomposition of product of matrices Given SVD(A) and SVD(B) and B is a diagonal matrix, is there a way or method to construct SVD(AB) ?
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 $A$.
This won't be a complete answer but maybe a step in the right direction.

Consider the equivalent problem of the SVD of $BA$, where $B$ is diagonal.  This is without loss of generality because if we know the SVD of $BA$, then we know the SVD of $(AB)^T=A^TB^T$, which is of the form you mention. We will assume that $B$ has non-zero diagonal elements, such that it has full rank.

Let the SVD of A be $USV^*$, where $U$ and $V$ are unitary, $S$ non-negative diagonal, and $V^*$ is the conjugate transpose of $V$.  Further, let $B=QR$ be the unique QR decomposition of $B$ obtained by the Gram-Schmidt method such that the diagonal elements of R are positive, where $Q$ is unitary and $R$ is upper triangular.  The positivity of the diagonal elements of $R$ will be useful in what follows.

Since $R$ is upper triangular, and $S$ is diagonal, $RS$ will also be diagonal. Namely, $RS$ will be a diagonal matrix formed of the products of the diagonal elements of $R$ with the corresponding elements of $S$. Since $R$ and $S$ both have positive diagonal elements, so does their product, which we will call $D$. Thus an SVD of $BA$ is given by:

$$BA=QDV^*$$