I understood pretty much everything besides the last statement here. What does the author mean when he says the image of s under the permutation $\pi_x$? I mean the permutation isn't a mapping right? How would it have an image for an element? Any kind explanation especially with an example will be very much appreciated.
Permutations are simply bijections from a set to itself, so $\pi_x$ is a mapping. Indeed, $\pi_x:S\rightarrow S$.
For example, let $G$ be a group with an index-$n$ subgroup $H$, and write $g_1H, \ldots, g_nH$ for the cosets. So we take $S=\{g_1H, \ldots, g_nH\}$. Take $x\in G$, and define $\pi_x$ by the rule $\pi_x(g_kH)=(xg_k)H=g_{k'}H\in S$ for some $g_{k'}$ (by results on cosets).
Defining each $\pi_x$ like this does define a homomorphism \begin{align*} \pi: G&\rightarrow \operatorname{Perm}(S)\\ x&\mapsto \pi_x \end{align*} and proving this is a nice exercise. But hopefully you see my point: $\pi_x$ is permutation of $S$, and so is clearly a map from $S\rightarrow S$.
The mapping $G\times S\rightarrow S$ here is simply $(x, g_kH)\mapsto (xg_kH)$.