A nilpotent group is one of the concepts that is most difficult to grasp, particularly for infinite groups. If $G$ is a finite nilpotent group then it is just a direct product of $p$-groups, and that's normally enough to satisfy yourself.
A soluble group $G$ of length $n$ is one where you take the commutator subgroup $G'$ and this has length $n-1$. That, together with the fact that the trivial group is soluble of length $0$, is enough to understand the class. In particular, if $N$ is a normal subgroup and both $N$ and $G/N$ are soluble, then $G$ is soluble.
Nilpotent is similar, but you need that the commutator for the normal subgroup is 'compatible' with the whole group. So instead of checking that $G'$ is soluble, i.e., $[G',G']<G'$ and so on, you want that the commutator works with one of the $G'$ replaced by the whole of $G$. So $H=[G',G]<G'$, and then $[H,G]<H$ and so on until you hit the trivial group.
So if we make the map $\mathrm{ad}_x:G\to G$ given by $y\mapsto [x,y]$ then this map is nilpotent, i.e., some power is the 'zero' map (i.e., sends every element to the identity). This is not true for soluble groups, e.g., $G=S_3$ with $x=(1,2)$. If you know ring theory, the analogue would be the difference between a subring and an ideal, where we want a compatibility between the multiplication on the whole of the ring, not just the subset, to move from a subring to an ideal.
Whereas $G$ is soluble if and only if both $G/N$ and $N$ are soluble, the same statement cannot hold for nilpotent groups because it gives no information that connects the commutator map for $N$ to that of $G$. A group $G$ is nilpotent if and only if both $G/N'$ and $N$ are nilpotent. That drop from $N$ to $N'$ gives us enough information to connect the two commutator structures.