Determining isomorphism classes of groups of the same order is difficult and largely an ad-hoc procedure. Looking for computable invariants is what you have to look for, such as centers, abelianizations and existence of elements of certain orders. Exactly which invariant is the "right" one isn't always easy to figure out, and takes a bit of experience and guesswork.
For direct products your (1) suffices for the "obvious" attempt to get a direct product, as if you have a semi-direct product expression then you already know the action $\phi$, which is easily verified to be (non)trivial. The tricky part happens when considering non-idenity maps, and even trickier if you don't already know a semidirect product expression. For this you may be interested in the direction extension theorem for finite groups (or groups with chain conditions):
If $G=H\times K$ is a finite group, and $H_0$ is a normal subgroup of $G$ such that $G/H_0\cong K$ and $H_0\cong H$, then $H_0$ is a direct factor of $G$. Equivalently, every short exact sequence $0\longrightarrow H\longrightarrow G\longrightarrow K\longrightarrow 0$ splits as a direct product.
You can in principle use this to verify that $G$ is not a direct product by exhibiting a case where the SES doesn't split (a semi-direct product "half-splits"). In your specific case, it suffice to find (or prove non-existence of) a group homomorphism $f\colon S_3\rtimes \mathbb Z_2\to S_3$ such that, if $g\colon S_3\to S_3\rtimes\mathbb Z_2$ is the obvious embedding, then $fg=\operatorname{id}_{S_3}$. I'll show how to find such an $f$ at the end of the answer. Normally there are many inequivalent SES's for a given $G,H,K$, but the theorem says in the direct product case they are all equivalent.
For part (2), see this MO Q&A and the comments therein. A counterexample to the general case is:
Pick $G$ dihedral of order 8, $A$ a Klein subgroup of order 4 and exponent 2, $C\subset A$ the center of $G$ (so $G/C$ is also a Klein group). Then $C$ is not part of a semidirect decomposition. – YCor
The answers consider some sufficient conditions where an iterated semidirect product may be "associative" in the sense you desire.
So consider $G=S_3\rtimes\mathbb Z_2$ with your action, and let $g\colon S_3\to G$ be the obvious injective homomorphism. As mentioned above, to show $G\cong S_3\times \mathbb Z_2$ it suffices to find $f\colon G\to S_3$ such that $f\circ g = \operatorname{id}_{S_3}$. Obviously $f$ must be surjective, so we need a normal subgroup of order 2 in $G$. The subgroups of order 2 are easily specified by the elements of order 2, which are clearly
\begin{array}{ccc}
((12),0) & ((13),0) & ((23),0)\\
((12),1) & ((13),1) & ((23),1).
\end{array}
The elements in the first row do not yield a normal subgroup since $S_3$ does not have normal subgroups of order $2$. The bottom left entry in the second row, however, does produce a normal subgroup. Indeed, it gives the center. The quotient thereof has image generated by the images of $((123),0)$ and $((12),0)$, which is clearly non-abelian and yields the desired $f$ in the obvious fashion. Explicitly,
\begin{align*}
f((123),0) &= (123)\\f((12),0) &= (12)\\f((12),1)) &= ()
\end{align*}
completely defines $f$. Alternatively, from the splitting condition we can also assert that the first two identities must hold for such an $f$ to exist, and then check a remaining generator (such as $((),1)$ or $((12),1)$) to see if there's a way to get a well-defined homomorphism. In this case, the action of the semi-direct product means we need the image of $((),1)$ to act by conjugation by $(12)$ in $S_3$, which will obviously happen if $((),1)\mapsto (12)$; equivalently, $((12),1)\mapsto ()$.
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.
Best Answer
The semi-direct product involves more than just two groups - it also requires, as you've correctly noticed, a group homomorphism.
And indeed, two different group homomorphisms may give us different semi-direct products, even non isomorphic semi-direct products. To see that note that the the trivial $\alpha(x)=id$ homomorphism gives us the standard direct product. You now just have to pick an example which is not a direct product. For example there are non-abelian semi-direct products of two abelian groups, e.g. there is a non-abelian semi-direct product of $\mathbb{Z}_8$ and $\mathbb{Z}_2$. In fact there are 4 non-isomorphic groups of order $16$, and each one is a semi-direct product of $\mathbb{Z}_8$ and $\mathbb{Z}_2$ (including the direct product).