Let $k$ be a field, and suppose $0 \to M \to N \to P \to 0$ is an exact sequence of $k$-vector spaces. From this answer, we see that for any $r \in \mathbb{Z}_{>0}$ the map $$\mathrm{Sym}^r(N) \to \mathrm{Sym}^r(P)$$ is surjective, and in fact we have an exact sequence $$\mathrm{Sym}^{r-1}(N) \otimes M \to \mathrm{Sym}^r(N) \to \mathrm{Sym}^r(P) \to 0$$ where the first map is given by $n_1 \cdots n_{r-1} \otimes m \mapsto n_1 \cdots n_{r-1} m$.
I suspect that we can extend this to get an exact sequence $$ \cdots \to \mathrm{Sym}^{r-3}(N) \otimes \wedge^3 M \to \mathrm{Sym}^{r-2}(N) \otimes \wedge^2 M \to \mathrm{Sym}^{r-1}(N) \otimes M \to \mathrm{Sym}^r(N) \to \mathrm{Sym}^r(P) \to 0$$ where the map $$\mathrm{Sym}^{r-i}(N) \otimes \wedge^i(M) \to \mathrm{Sym}^{r-i+1}(N) \otimes \wedge^{i-1}(M)$$ is given by $$n_1\cdots n_{r-i} \otimes (m_1 \wedge \cdots \wedge m_i) \mapsto \sum_{j=1}^i (-1)^{j+1} n_1\cdots n_{r-i} m_j \otimes (m_1 \wedge \cdots \hat{m_j} \cdots \wedge m_i)$$ where $\hat{m_j}$ means we omit this factor, but I’m having trouble proving or disproving this.
If this is indeed true, I’m guessing it should also be true for free $A$-modules (where $A$ is an arbitrary commutative ring with $1$), but how about for arbitrary $A$-modules?