$[G,G] = \gamma_2$ is not finitely generated. But it is the normal closure of the set $\{ [x,y],[x,z],[y,z]\}$ of the commutators of the three generators.
In the quotient group $\gamma_2/\gamma_3$, all conjugates of these generators have the same images. For example $[x,y]^z\gamma_3 = [x,y]\gamma_3$, because $[x,y]^{-z}[x,y] = [z,[x,y]] \in \gamma_3$.
So $\gamma_2/\gamma_3 = \langle [x,y]\gamma_3,[x,z]\gamma_3,[y,z]\gamma_3 \rangle$.
The first of these three generators has infinite order, but $z^2=1$ implies that the second and third have order $2$, so $\gamma_2/\gamma_3 \cong Z \oplus Z/(2Z) \oplus Z/(2Z)$ (I have been lazy and written $Z$ instead of ${\mathbb Z}$.)
It is possible to compute further factors $\gamma_i/\gamma_{i+1}$ but it gets much more difficult as $i$ increases.
Consider $G=V\times D_8$ (direct product of Klein four group with the group of the square) of order 32. The lower central series is $1<1\times Z(D_8)<G$. The upper series is $1<V\times Z(D_8)<G$. And three additional central series are given by $1<H\times Z(D_8)<G$ where $H$ is any of the three order 2 subgroups of $V$. Thus $V\times D_8$ is an example of what you are looking for.
Note that I have answered the strict version of this problem, where the additional central series have the same lengths as the upper and lower series. If they are allowed to be longer, the problem is even easier.
As for why the upper series' terms always contain the corresponding lower series, this is a standard result you can find in many textbooks. The essential idea is to start at the top, noting that $G'=\gamma_2(G)$ is the smallest normal subgroup with abelian quotient, so that $Z_{c-1}(G)\ge \gamma_2(G)$. And continue by an easy induction.
Best Answer
A finite group $G$ is nilpotent if and only if the lower central series ends at $1$, if and only if the upper central series ands at $G$. Since $S_4$ has trivial center, it cannot be nilpotent. Consequently, its lower central series cannot and at $1$ - and it doesn't, as you have shown.