Let $G/H=K$. I think I am right in saying that if $K\cong K'\trianglelefteq G$ and is such that $K'H=G$ and $K'\cap H=1$ then we have $H\times K'=G$. If we relax the normality requirement of $K'$ to simply $K'\le G$ we have $H\rtimes K'=G$. I think there is no need for Zappa-Szep products here.
Now there will be situations where I don't know whether the image is isomorphic to a subgroup or a normal subgroup, or if $K'\cap H=1$, or if $K'H=G$.
It has been pointed out in the comments that there is no general result for finding out the information about the image and kernel required to make a direct or semidirect product. What are some special cases? Finite abelian groups have been mentioned.
In general $G/H$ is not isomorphic to a subgroup of $G$.
For example, $G = (\mathbb Z,+)$ the group of integers, $H = (2\mathbb Z,+)$ the subgroup of even integers. The quotient $G/H$ has order $2$, so it cannot be a subroup of $G$.