Yes, that's correct. This is just the algebra homomorphism $\mathrm{Sym}_A(M) \to \mathrm{Sym}_A(0)$ induced by the module homomorphism $M \to 0$. You can also write down the addition on $E/X$, this corresponds to the algebra homomorphism $\mathrm{Sym}_A(M) \to \mathrm{Sym}_A(M) \otimes_A \mathrm{Sym}_A(M) = \mathrm{Sym}_A(M \oplus M)$ which is induced by the evident module homomorphism $M \to M \oplus M$.