The screenshot is taken from Zieschang's book Knots and describes a way to find the longitude $l$ of a knot. Recall that the peripheral system is a choice of meridian $m$ and longitude $l$ such that $m$ bounds a disk in the closed regular neighbourhood of the knot and $l$ is killed in the first homology group of the knot complement.
In this picture, with the chosen longitude $l=s_4^{-1}s_5^{-1}s_2^{-1}s_1^{-1}s_3^{-1}s_1^{5}$, does $l$ commute with each $s_i$? Dose $(s_i,l)$ form a peripheral system for each $i$? I know from Wirtinger presentation, each $s_i$ is a meridian. But I feel confused about the peripheral systems.
Let me start by saying that element $s_i$ is a meridian, then if you look at the figure you can see that $s_4^{-1}s_{5}^{-1}s_2^{-1}s_3^{-1}$ is a "parallel" copy of the knot $K$. If we compute its homology class we see that it is not zero and. Thus, we multiply the "parallel" copy of $K$ by $s_1^5$ to make it zero.
Consider now $S^3 \setminus \eta(K) = X(K)$, with $\eta(K)$ a tubular neighborhood of $K$. The curves $\{s_1, \cdots, s_5,l\}$ are curves (actually slopes) in the torus $\partial X(K)$. This answers your first question: since $s_1,\cdots,s_5$ and $l$ can be seen as curves on a torus, then they are inside the image of the map $i_*:\pi_1(\partial X(k)) \to \pi_1(X(K))$ with $i$ the inclusion. Since the torus has an abelian fundamental group, $s_i$ commutes with $l$.
For your second question: the peripheral system of a knot $K \subset S^3$ is the subgroup of $\pi_1 (X(K))$ generated by a meridian and the longitude. Since all the meridians $s_i$ are conjugated (see the Wirtinger relations) then the peripheral system (or subgroup) is generated by $\{s_i,l\}$ for all $i=1,\cdots,5$.
Note that I used $K=5_2$ only for defining the longitude $l$ and then you can generalize this to every knot $K$.