Symmetric powers of modules Let
$$0 \rightarrow M \rightarrow N \rightarrow P \rightarrow 0$$
be an exact sequence of $A$-modules, where $A$ is a commutative ring with unity.
Then for each natural number $i$ one can look at the induced maps
$$\mathrm{Sym}^i(N) \rightarrow \mathrm{Sym}^i(P)$$
and 
$$\mathrm{Sym}^i(M) \rightarrow \mathrm{Sym}^i(N).$$
The first question is, if they are still surjective, resp. injective, i.e. is $\mathrm{Sym}^i$ an exact functor?
The second question is: what are the kernel, resp. cokernel of these maps?
 A: Surjectivity: yes
The morphism  $Sym^i(N) \rightarrow Sym^i(P)$ is surjective.
 Its kernel is generated by products  of the form $mn_2n_3...n_i$ with $ m \in M$   and $n_k\in N$
Injectivity: no
The morphisms $Sym^i(N) \rightarrow Sym^i(P)$   needn't be injective just because  $f:M\to N$  is injective.      
Let me give an example where $f:M\to N$ is injective but $S^2(f):S^2(M) \to S^2(N)$ is not.
Take $A=\mathbb Z/(4) , N=A, M=2A$ and $f: M\to N$ is the inclusion .
The crucial point is that $S^2(N)=N\otimes_A  N  $  and $S^2(M)=M\otimes_A  M$ because these modules are generated by a single element (Check it on the definition of symmetric product!).
So it suffices to show that  $f^{\otimes 2} : M\otimes_A  M \to N\otimes_A  N:2a\otimes 2b\mapsto 2a\otimes 2b=0 $ is not injective.
Since this map is zero it suffices to show that $M\otimes_A  M\neq 0$.
But this is clear because $M\simeq A/(2)$ as $A$-modules and $A/(2)\otimes_AA/(2) \simeq A/(2)$ as $A$-modules.
Edit
Since  Cyril asks, let me indicate the plan of the proof that the kernel of $Sym^i(N) \rightarrow Sym^i(P)$ is generated as an $A$-module by the $mn_2n_3...n_i$ with $m \in M$ and $n_k\in N$.
It suffices  to prove the  analogous result that the kernel of  $\otimes^iN \rightarrow \otimes^i P$ is generated by the $n_1\otimes n_2\otimes...\otimes n_i$ with one of the $n_k$ lying in $M$, and then take suitable quotients.
The assertion on the kernel of the $i$-th tensor product is based on the right-exactness of the tensor product. 
Full details can be found in Bourbaki , Algebra, Chapter  III, §6.2, Proposition 4, page 499.
A: $\mathrm{Sym}$ 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
