In the definition of the outer semidirect product it is usually said that there is a group homomorphism $\phi: H \to Aut(N)$ from the elements of $H$ to the automorphism group of $N$. I get that. I also get that $\phi_h$ is the specific automorphism of $N$ defined by some specific element of $H$. What I'm unclear on is how to interpret $\phi_h(n)$.
I think it means that it means we should map $n$ to $m$, where $m$ is the element mapped to $n$ by that automorphism but I'm not confident of that.
I believe it is worth starting with the interior direct product, that is, you have a group $G$, and two subgroups $H, N$ such that $N$ is normal in $G$ and $G = H N$.
Then $$\phi_h(n) = h n h^{-1}.$$
Once you have defined the product in the exterior semidirect product you are considering, the same formula will hold in there.