Let $G$ be a group with $N \unlhd G$ such that there is $K \leq G$ with $KN = G$ and $K \cap N = 1$.
It is well known that in this case $$G \cong N \rtimes_\phi K$$
where $\phi: K \to Aut(N): k \mapsto (\phi_k: N \to N: n \mapsto knk^{-1})$
In exercises on classifying groups, we look for subgroups $K, H$ as above and then we conclude that $G \cong N \rtimes H$ for some $\phi: K \to Aut(N)$. We then proceed to find out what possibilities there are for such a map.
Why don't we just say that it is the $\phi$ as above? Is it because we don't want our description of $\phi$ to depend on conjugation in $G$?
Because otherwise, $G \cong N \rtimes K$ isn't useful because $\phi$ depends on how we calculate conjugation in $G$ and we want to describe $G$ independently of its subgroups?
In general, $\phi$ is just any group homomorphism from $K$ to $Aut(N)$. If we have $G=KN$, $K\cap N=1$, we have an inner semidirect product, with the $\phi$ you have given.
"Why don't we just say that it is the $\phi$ as above?" Because there are also outer semidirect products, $G\cong N\rtimes_{\phi} K$ with an arbitrary homomorphism $\phi\colon K\rightarrow Aut(N)$. This is equivalent to the fact that there is a split short exact sequence of groups $$ 1\rightarrow N\rightarrow G \rightarrow K\rightarrow 1. $$