The "standard" proof:
Consider, for each $i$, the subgroups $G_iN/N$ of $G/N$. It is straight-forward to show that $G_iN$ is normal in $G_{i+1}N$ (and thus $G_iN/N$ is normal in $G_{i+1}N/N$).
Now $\dfrac{G_{i+1}N/N}{G_iN/N} \cong G_{i+1}N/G_iN$.
Taking any $x,y \in G_{i+1}$, and $n,n' \in N$, we see that:
$[xn(G_iN),yn'(G_iN)] = [xn,yn']G_iN$, and since $N \lhd G$, $[xn,yn'] \in [x,y]N$, so that:
$[xn,yn']G_iN = [xn,yn']NG_i = ([xn,yn']N)G_i = $
$([x,y]N)G_i = [x,y]NG_i = [x,y]G_iN = ([x,y]G_i)N$.
Since $G_{i+1}/G_i$ is abelian, $[x,y]G_i = G_i$ so $[xn,yn']G_iN = G_iN$,so that
$G_{i+1}N/G_iN$ is abelian, as desired.
(here, $[x,y] = xyx^{-1}y^{-1}$ and we are using the fact that for a group $G$, $G$ is abelian iff $[x,y] = e$ for all $x,y \in G$).
It should be noted that some of the quotients $G_{i+1}N/G_iN$ may be trivial, resulting in a shorter subnormal series for $G/N$.
Forget about the actual construction of the semidirect product for now. I argue that the semidirect product is important because it arises naturally and beautifully in many areas of mathematics. I will list below many examples, and I urge you to find a few that interest you and look at them in detail.
Before doing that let me just give the following extra motivation: say you have a group $G$, and you find two subgroups $H,K$ such that every element of $G$ can be uniquely written as a product $hk$ for $h \in H$, $k \in K$. In other words, you have a set-theoretic bijection between $G$ and $H \times K$. Then certainly you'd want to understand $G$ by studying its smaller components $H$ and $K$. One way to achieve this would be to find a suitable group structure on $H \times K$ intertwining the structures of $H$ and $K$ such that the above bijection becomes a group isomorphism. This can be done; however doing things in this generality becomes rapidly tedious. If instead we restrict our attention to such decompositions with $H$ normal in $G$, the problem becomes much more manageable. In this case we have what we call a split exact sequence $$1 \to H \to G \to K \to 1,$$
and $G$ is called a semidirect product of $H$ and $K$. An existence and uniqueness theorem gives us all the possible semidirect products one can obtain from $H$ and $K$ through the group of homomorphisms from $K$ to $\operatorname{Aut}H$. Note that $K = \mathbb{Z}/2$ appears often in practice, because this guarantees normality of $H$. Now here are some examples:
The symmetric group $S_n = A_n \rtimes \mathbb{Z}/2$. The exact sequence is $$1 \to A_n \to S_n \xrightarrow{\mathit{sign}} \mathbb{Z}/2 \to 1.$$
The dihedral group $D_n = \mathbb{Z}/n \rtimes \mathbb{Z}/2$. The exact sequence is $$1 \to \mathbb{Z}/n \to D_n \xrightarrow{\mathit{det}} \mathbb{Z}/2 \to 1.$$
The infinite dihedral group $D_\infty = \mathbb{Z} \rtimes \mathbb{Z}/2$. The exact sequence depends on your explicit construction. You may take $$1 \to \mathbb{Z} \to \mathbb{Z}/2 * \mathbb{Z}/2 \to \mathbb{Z}/2 \to 1$$ or $$1 \to \mathbb{Z} \to A(1,\mathbb{Z}) \to \mathbb{Z}/2 \to 1,$$ where $A(1,\mathbb{Z})$ is the group of affine transformations of the form $x \mapsto ax + b$, where $a \in \{ \pm 1 \} \cong \mathbb{Z}/2$ and $b \in \mathbb{Z}$.
Many matrix groups, thanks to the determinant map. For example, $G = \operatorname{GL}(n,\mathbb{F})$, $O(n,\mathbb{F})$ and $U(n)$ have respective subgroups $H = \operatorname{SL}(n,\mathbb{F}),\operatorname{SO}(n,\mathbb{F}),\operatorname{SU}(n)$ and $K = \mathbb{F}^\times,\mathbb{Z}/2,U(1)$.
The fundamental group of the Klein bottle is $G = \langle x,y \mid xyx = y \rangle$. This is just the nontrivial semidirect product of $\mathbb{Z}$ with itself. Interestingly, the other (trivial) semidirect product $\mathbb{Z}^2$ is the fundamental group of the other closed surface of Euler characteristic $0$, namely the torus.
The affine group $A(n,\mathbb{F}) = \mathbb{F}^n \rtimes \operatorname{GL}(n,\mathbb{F})$. Its elements are transformations $\mathbb{F}^n \to \mathbb{F}^n$ of the form $x \mapsto Ax + b$, with $A$ an invertible matrix and $b$ a translation vector. The exact sequence is $$1 \to \mathbb{F}^n \to A(n,\mathbb{F}) \xrightarrow{f} \operatorname{GL}(n,\mathbb{F}) \to 1$$
where $f$ forgets the affine structure (the translation part).
The hyperoctahedral group $O(n,\mathbb{Z})$ is the group of signed permutation matrices. We have two decompositions $O(n,\mathbb{Z}) \cong \operatorname{SO}(n,\mathbb{Z}) \rtimes \mathbb{Z}/2$ and $O(n,\mathbb{Z}) \cong (\mathbb{Z}/2)^n \rtimes S_n$. In the corresponding exact sequences the surjective map is respectively the determinant homomorphism and the "forget all the signs" homomorphism.
Best Answer
If I've understood the question correctly, then no. There are soluble but non-supersoluble groups that cannot be expressed as a semi-direct product. I think the smallest example is the binary octahedral group of order $48$. It has $3$ proper non-trivial normal subgroups, and one can check (via computer) that none of them is complemented.