Can I write $$|| V D_1 D_2 V^T -Q D_1 D_2 Q^T||_F \le f(D_1,D_2)_F || V D_2 V^T -Q D_2 Q^T||_F $$
for some $f$? I have that $V$ and $Q$ are orthonormal matrices, $D_1$ and $D_2$ are diagonal matrices. I don't think Cauchy-Schwarz covers it. I don't need this precise inequality, I just want a bound as a function of the two terms shown on the right-hand side.
No such $f$ exists. Consider the case where $D_2=I$ and $V,Q$ are orthogonal matrices. Then the RHS of your inequality is zero. Hence the LHS is necessarily zero, i.e., the inequality implies that $VD_1V^T=QD_1Q^T$ for every diagonal matrix $D_1$ and orthogonal matrices $V,Q$, which is clearly untrue.
However, if we put $X=VD_1V^T,\,Y=VD_2V^T,\,Z=QD_1Q^T$ and $W=QD_2Q^T$, we do have $$ \begin{aligned} &\|VD_1D_2V^T-QD_1D_2Q^T\|_F\\ &=\|XY-ZW\|_F\\ &=\|XY-XW+XW-ZW\|_F\\ &\le\|XY-XW\|_F+\|XW-ZW\|_F\\ &=\|X(Y-W)\|_F+\|(X-Z)W\|_F\\ &\le\|X\|_F\|Y-W\|_F+\|X-Z\|_F\|W\|_F\\ &\le\|D_1\|_F\|VD_2V^T-QD_2Q^T\|_F+\|VD_1V^T-QD_1Q^T\|_F\|D_2\|_F. \end{aligned} $$