$I=(i_1,...,i_k)$ denotes an ordered $k$-tuple of indices.
Given $\sigma\in S_k$, define $\sigma(I)=(i_{\sigma^{-1}(1)},...,i_{\sigma^{-1}(k)})$
Let $\sigma,\tau\in S_k$. Then $\sigma(\tau(I))=(\sigma\tau)(I)$
Proof:
(1) Let $J=(j_1,...,j_k)=\tau(I)$ and $K=(k_1,...,k_k)=\sigma(J)$. Then $j_r=i_{\tau^{-1}(r)}$ and $k_r=j_{\sigma^{-1}(r)}$ such that
(2) $k_r=i_{\tau^{-1}(\sigma^{-1}(r))}=i_{(\sigma\tau)^{-1}(r)}$
(3) Therefore $K$ which we defined to be $\sigma(\tau(I))$, is also given by $(\sigma\tau)(I)$
I understand the general argument of the proof but I dont understand the step in part (2) that $k_r=i_{\tau^{-1}(\sigma^{-1}(r))}$
In (1), the index $r$ is arbitrary, so if you accept (1), then you have: $$ k_r = j_{\sigma^{-1}(r)} $$ and by substituting $\sigma^{-1}(r)$ in place of $r$, $$ j_{\sigma^{-1}(r)} = i_{\tau^{-1}(\sigma^{-1}(r))} $$