Let $G$ be a group and $H$ be a subgroup of $G$. Suppose $G$ is an extension of $H$ such that $G = N \rtimes H.$ Consider a normal subgroup $K$ of $G,$ which is also an extension of $H.$
Clearly, $K = (N \cap K) \rtimes H.$
I was trying to know what should be the condition on such a subgroup $K$ of $G$ so that it itself is an extension of $H.$ Also, can we say that $N$ is a direct product of $N \cap K$ with some central subgroup of $G$?
Clearly, the first one that comes to mind is $H \subseteq K.$ Other one is $\frac{G}{N} \cong \frac{K}{N \cap K}$ and so $\left|\frac{G}{N}\right| = \left|\frac{K}{N \cap K}\right|.$
Is there some other important information about $K$ that I am missing?
The only required condition is $H\leq K$.
Suppose $H\leq K$, then set $M:=N\cap K$. Then:
Hence, we have the three conditions for a semidirect product, so can conclude that $K=M\rtimes H$.
(There is a warning here: we are talking about a fixed subgroup $H\leq G$, so $K$ may contain an isomorphic copy of $H$ but not split like this. For example, consider $\mathbb{Z}_2\times\mathbb{Z}_4=H\times N$. Taking $K$ to be $N=\mathbb{Z}_4$, this contains an isomorphic copy of $H=\mathbb{Z}_2$, but does not split as a (semi-)direct product.)
The question also asks if we necessarily have $N=M\times Z$ for some central subgroup $Z$ of $G$. This is not necessarily the case. For example, suppose we are given a semidirect product $P\rtimes H$ with $P$ non-abelian, then we can form the semidirect product $(P\times P)\rtimes H$ with $P$ non-abelian, where $(p_1, p_2)^h=(p_1^h, p_2^h)$ (i.e. extend the action of $H$ on $P$ to $P\times P$ in the natural way). Then using the notation from the question, take $N=P\times P$ and let $K=(1\times P)\rtimes H$. Then $M=N\cap (P\times P)=1\times P$ is not central in $G$. I am sure similar constructions can produce examples of $M$ and $N$ with even less "commutativity".