Why is trivial intersection of groups $N$ and $H$ not required in the definition of outer semi-direct products?

Is $\mathcal{N}\cap\mathcal{H} =\{(1_N,1_H)\}$ by construction?

Yes. $\mathcal{N}$ is equal to $N\times\{1_H\}$, and $\mathcal{H}$ is equal to $\{1_N\}\times H$. The intersection of these two sets is $\{e\}$, where $e=(1_N,1_H)$ is the identity of $N\rtimes_\varphi H$.

The reason we can't stipulate $N\cap H=\{1\}$ in advance is that "$1$" doesn't refer to anything concrete, at least not before you make further identifications. For example $N$ could be a set of integers and $H$ could be a set of automorphisms of some Riemannian manifold. How would you intersect them to get something meaningful? You have to make an identification first. This identification is almost exactly what $N\rtimes_\varphi H$ accomplishes.

$\mathcal N = \{(n,1_H) \mid n \in N\}$ and $\mathcal H = \{(1_N,h) \mid h \in H\}$. It follows $(n,h) \in \mathcal N \cap \mathcal H$ if and only if $n=1_N$ and $h=1_H$. So yes, $\mathcal N \cap \mathcal H = \{(1_N,1_H)\}$.