$\mathrm{Sym}$ is not exact: just think about dimension. (I will work over some field.) In your set-up, if $M,N,P$ have dimensions $m,n,p$ then $n=m+p$. The symmetric square of a vector space of dimension $k$ has dimension $k \choose 2$, and in general $n \choose 2$ will be different from ${m \choose 2} + {p \choose 2}$. (They are always different unless $m=p=1$.)
$\mathrm{Sym}$ does preserve surjections. I will write monomials in the symmetric power of $P$ as $p_1 \vee \cdots \vee p_r$; such elements span the $r$th symmetric power. If $n_i$ is a preimage of $p_i$ under your surjection, $n_1\vee\cdots\vee n_r$ is a preimage of $p_1 \vee \cdots \vee p_r$ under the induced map on the symmetric power.
However $\mathrm{Sym}$ is not right-exact. Regard $M$ as a subspace of $N$, consider an element of the symmetric square of the form $m \vee n$ where $m \in M, n \in N \setminus M$. This is in the kernel of the induced map $\operatorname{Sym}^2 (N) \to \operatorname{Sym}^2 (P)$ but not in the image of $\operatorname{Sym}^2(M) \to \operatorname{Sym}^2(N)$. Thus
$ \operatorname{Sym}^2(M) \to \operatorname{Sym}^2(N) \to \operatorname{Sym}^2(P) \to 0$ is not exact. (You seem to have the wrong idea of what "exact" means: e.g. for a covariant functor $F$ to be right-exact it must send exact sequences $A\to B\to C \to 0$ to exact sequences $FA \to FB \to FC \to 0$ which is stronger than simply preserving surjections.)
$\mathrm{Sym}$ is also not a linear functor, that is,
$ \operatorname{Sym}^2 : \hom(V,W) \to \hom(\operatorname{Sym}^2(V),\operatorname{Sym}^2(W)) $
is not a linear map. This causes technicalities when trying to define the derived functors, but that's not to say it is impossible. See http://arxiv.org/abs/0911.0638