When is a group $G$ isomorphic to $(G/N) \times N$? Let $G$ denote a group and $N$ denote a normal subgroup thereof. Then there's always a bijection $G \rightarrow (G/N) \times N$, owing to the fact that $G$ and $(G/N) \times N$ have the same cardinality. However, these groups usually aren't isomorphic. For example, if $G = \mathbb{Z}/4$ and $N = \mathbb{Z}/2$, then $(G/N) \times N$ is $\mathbb{Z}/2 \times \mathbb{Z}/2$, which is not isomorphc to $\mathbb{Z}/4.$
Sometimes however they're isomorphic. For example, if $G$ is a finitely-generated abelian group and $N$ is its subgroup of torsion elements, then a weak form of the structure theorem for finitely-generated abelian groups says that $G$ is isomorphic to $(G/N) \times N$.

Question. For which normal subgroups $N$ is this true?

 A: A trivial answer to your question that the normal subgroup $N$ splits off as a direct factor of $G$:
$$
G=N\times Q$$
for some subgroup $Q< G$. (I.e. there exists another normal subgroup $Q< G$ such that $Q\cap N=\{1\}$, $[N,Q]=\{1\}$ and $G=NQ$.) Whether you are looking for anything beyond this statement is unclear to me. 
Edit. It is clear that if $G=G_1\times G_2$ then $G/G_1\cong G_2$, the isomorphism is given by the map $G_1\times g_2 \mapsto g_2$. 
A: As Arthur mention in a comment, whenever there is a group homomorphism $p:G\to N$ whose restriction to $N$ is the identity, then $G$ is isomorphic to $N\ltimes \ker(p)$, where the action of $N$ on $\ker(p)$ is given by conjugation in $G$. But if $N$ is assumed to be normal, then this action is actually trivial, and thus $G$ is actually isomorphic to $N\times\ker(p)$. Indeed, in that case, for every $n\in N$ and $k\in \ker(p)$ we have $nkn^{-1}k^{-1}\in N$, and thus $$nkn^{-1}k^{-1}=p(nkn^{-1}k^{-1})=nn^{-1}=1,$$
which means that $nkn^{-1}=k$.
So a possible answer to your question is that there is an isomorphism if and only if $N$ is a retract of $G$.
