Why does semidirect product give a complete classification of small groups The key theorem involved is:
Suppose $G$ is a group with subgroups $H$ and $k$ such that $H$ is normal in $G$ and $H \cap K=1$. Let $\phi: K \to Aut(H)$ be the homomorphism defined by mapping $k \in K$, TO THE automorphism by conjugation by $k$. Then if $G = HK$, we have $G$ is the semidirect product of $H$ and $K$ with respect to $\phi$.
I am just confused about how to use this to classify groups of small orders. Say for instance I have a noncommutative group with order $pq$, $p$ prime, $q$ prime, $p < q$. Then Let $H$ be the $q-$sylow subgroup of $G$, $K$ be a $p-$sylow subgroup of $G$. Then $H \cap K = 1$, and $H$ is normal in $G$. It is also not hard to show that $G = HK$. Then by the theorem shouldn't we naturally have that $G$ is the semidirect product of $H$ and $K$ with respect to the homomorphism induced by conjugation?
However, all illustrations on this example start by describing all the possible homomorphisms from $Z_p \to Z_{q-1}$. I am totally lost here, why should we care about homomorphisms other than the one induced by the conjugation?

The semidirect product can be defined with respect to any homomorphism $\phi$ from $K$ to the group of automorphisms of $H$. The trivial homomorphism (everything goes to the identity) gives the direct product, which always exists. The automorphisms of $H$ in the image of $K$ need not all be inner automorphisms, and even if the image of the homomorphism consists of inner automorphisms, $\phi$ need not be the homomorphism defined by conjugation.
It therefore has to be proved that, where non-abelian groups of order $pq$ exist, they are all isomorphic. In this simple case this is not hard to do.
