$\begin{aligned}
&\phi(ss') = \\
&=\phi((n, k) \cdot (n', k')) \\
&\text{definition of semidirect product via conjugation:} \\
&= \phi((n {n'}^k, kk')) \\
&\text{definition of $\phi$:} \\
&= n n'^{k} kk' \\
&\text{definition of $n'^k = k n' k^{-1}$:} \\
&= n k n'k^{-1} k k' \\
&= n k n' k' \\
&= \phi(s) \phi(s')
\end{aligned}$

So, $\phi$ really is a homomorphism from the external description (given in terms of the conjugation)
and the internal description (given in terms of the multiplication).
We can also go the other direction, to start from the internal definition and get to the conjugation.
Let $g \equiv nk$ and $g' \equiv n'k'$. We want to multiply them, and show that the multiplication
gives us some other term of the form $NK$:
$\begin{aligned}
gg' \\
&= (n k) (n' k') \\
&= n k n' k' \\
&= \text{insert $k^{-1}k$: } \\
&= n k n' k^{-1} k k' \\
&= n (k n' k^{-1}) k k' \\
&= \text{$N$ is normal, so $k n' k^{-1}$ is some other element $n'' \in N$:} \\
&= n n'' k k' \\
&= N K
\end{aligned}$

So, the collection of elements of the form $NK$ in $G$ is closed. We can check that the other properties
hold as well.