Suppose we have morphisms $f : A' \to A$, $g : A \to A''$, $r : A \to A'$, $s : A'' \to A$ in an abelian category. We have a split exact sequence if and only if these equations hold:
\begin{align}
r \circ f & = \textrm{id}_{A'} &
g \circ f & = 0 \\
r \circ s & = 0 &
g \circ s & = \textrm{id}_{A''} \\
\end{align}
$$f \circ r + s \circ g = \textrm{id}_A$$
These are precisely the same equations required to make $A$ into the direct sum $A' \oplus A''$, and in particular
$$0 \longrightarrow A' \longrightarrow A \longrightarrow A'' \longrightarrow 0$$
is an exact sequence. It is clear that $f$ is monic and $g$ is epic, and $g \circ f = 0$ implies $\operatorname{im} f \subseteq \ker g$; we only need to check that $\operatorname{im} f \supseteq \ker g$ now. So suppose $g \circ x = 0$ for some $x : X \to A$. Then, $s \circ g \circ x = 0$ as well, so
$$x = (f \circ r + s \circ g) \circ x = f \circ r \circ x$$
and therefore $\ker g \subseteq \operatorname{im} f$ indeed.
For the first part, the general definition is that a sequence $\cdots \to L \xrightarrow{f} M \xrightarrow{g} N \to \cdots $ is exact at $M$ if and only if the image of $f$ is the kernel of $g$.
A short exact sequence is an sequence consisting of five terms whose endpoints are zero, and is exact at every (internal) point. i.e.
$$ 0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0 $$
is exact at $A$, at $B$ and at $C$. Exactness at $A$ means that the kernel of $f$ is the image of the zero map: i.e. it is $\{ 0 \}$, so exactness at $A$ is equivalent to $f$ being injective. Similarly for exactness at $C$.
Usually it doesn't even make sense to say $B/A$, because $A$ usually isn't a submodule of $B$! However, there is equivalence in the sense that there is an isomorphism of short exact sequences:
$$ \begin{matrix}
0 &\to & A &\xrightarrow{f}& B &\xrightarrow{g}& C &\to& 0
\\ & & \ \downarrow^f & & \ \downarrow^1 & & \downarrow
\\ 0 &\to & f(A) &\xrightarrow{i}& B &\xrightarrow{\pi}& B / f(A) &\to& 0
\end{matrix}
$$
where both rows are short exact sequences, $i$ and $\pi$ are the inclusion and projection morphisms, all three vertical morphisms are isomorphisms, and the diagram is commutative. (meaning, e.g., if you take all paths from the top $A$ to the bottom $B$ and compose the arrows, you get the same result: $i \circ f = 1 \circ f$)
One can view that the situation as per the definition of quotient module is somewhat deficient, and the idea of a short exact sequence is what one really should be thinking about.
Best Answer
Hint:
Look at the sequence
$$ 0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2 \to 0 $$
as $\mathbb{Z}/2$-modules. Here the action on each module is by $x \mapsto -x$ (note this makes the action on $\mathbb{Z}/2$ trivial). Notice this means the action commutes with each of the module homs as well, so this really is a sequence of $\mathbb{Z}/2$-modules.
Now take $\mathbb{Z}/2$ invariants. Do you see what you get in each case? I'll include the answer under the fold, but you should really try to compute it yourself first!
As an aside, one can show the $G$-invariant functor is naturally isomorphic to the functor $\text{Hom}_{\mathbb{Z}G}(\mathbb{Z}, -)$. This might give some intuition for why this functor is left exact, but not exact in general. In fact, just like we can measure the failure of exactness of $\text{Hom}$ by looking at cohomology, there is a notion of group cohomology which measures the failure of exactness for this functor.
I hope this helps ^_^