Say $G/N=Q$ and $G,N,Q$ are all groups.
$N$ is the normal subgroup and $Q$ is the quotient group of $G$.
Let us say we require group homormophism for every possible map below. Is it true that
We can write every element in $g \in G$ as $g=(n,q)$ such that $n \in N$ and $q \in Q$? Here we do not need to ask to have a 1-to-1 bijective between $g$ and $(n,q)$.
Is that $g$ to $(n,q)$ surjective? In the sense that we can map a $g \in G$ to $(n,q) \in (N,Q)$? So $$G \to N \times Q$$ is also surjective?
When do we have injective between $G \to N \times Q$? (What criteria?)
When do we have bijective between $G \to N \times Q$? (What criteria?)
There is a bijection $G\to N\times Q$ as follows:
I'll take $Q$ to be the set of right cosets of $N$. For each $q\in Q$ fix a representative $g_q\in G$ which maps to $q$ in the quotient map $G\to Q$. The bijection is $x\mapsto(xg_{Nx}^{-1},Nx)$. This has inverse $(n,q)\mapsto ng_q$ so is bijective.
Rather trivially, the above is a homomorphism if and only if $G\cong N\times Q$.
As for when an injective homomorphism $G\to N\times Q$ exists, I don't believe there's any general criteria, other than the trivial case $G\cong N\times Q$ - it's certainly not possible in general, for example $G=\mathbb{Z}/4\mathbb{Z}$ has normal subgroup and quotient $N\cong Q\cong\mathbb{Z}/2\mathbb{Z}$, but $G\not\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$.
It is however possible to find infinite classes of non-trivial examples where it is possible, like $G=\mathbb{Z}$ and $N=n\mathbb{Z}$ for any integer $n$, with injective map $x\mapsto (nx,0)$