Let $A$ be a commutative ring with $f,g\in A[x]$ monics. Consider the $A$-linear endomorphism $\mu_g^{(f)}\in \mathrm{End}_A\tfrac{A[x]}{\langle f\rangle}$ given by multiplication by $g$.

For monics $f_1,f_2\in A[x]$, how to directly prove that $\det \mu_g^{(f_1f_2)}=\det\mu_g^{(f_1)}\det\mu_g^{(f_2)}$?

Writing down the matrix representation of $\mu_g^{(f)}$ w.r.t the monomial basis $1,x,\dots ,x^{\deg f-1}$ is messy and I am unable to see anything through it. On the other hand, $\mu_g^{(f)}=g(\mu_x^{(f)})$, and the matrix representation of $\mu_x^{(f)}$ w.r.t the monomial basis is the companion matrix of $f$, but again I see nothing smart to say about polynomial functions of a companion matrix.