I have expanded the expression in (3.7) by giving a number to the permutation. $$A(A(f)\otimes g)=\Bigg\{\text{sgn}(\sigma_{1})\text{sgn}({\tau_{1}})\sigma_{1}\tau_{1}f\otimes g+\text{sgn}(\sigma_{1})\text{sgn}(\tau_{2})\sigma_{1}\tau_{2}f\otimes g+...+\text{sgn}(\sigma_{1})\text{sgn}(\tau_{k!})\sigma_{1}\tau_{k!}f\otimes g\Bigg\}+\Bigg\{\text{sgn}(\sigma_{2})\text{sgn}({\tau_{1}})\sigma_{2}\tau_{1}f\otimes g+\text{sgn}(\sigma_{2})\text{sgn}(\tau_{2})\sigma_{2}\tau_{2}f\otimes g+...+\text{sgn}(\sigma_{2})\text{sgn}(\tau_{k!})\sigma_{2}\tau_{k!}f\otimes g\Bigg\}+...+\Bigg\{\text{sgn}(\sigma_{(k+l)!})\text{sgn}({\tau_{1}})\sigma_{(k+l)!}\tau_{1}f\otimes g+\text{sgn}(\sigma_{(k+l)!})\text{sgn}(\tau_{2})\sigma_{(k+l)!}\tau_{2}f\otimes g+...+\text{sgn}(\sigma_{(k+l)!})\text{sgn}(\tau_{k!})\sigma_{(k+l)!}\tau_{k!}f\otimes g\Bigg\}$$. I know that $$\sigma_i\tau_j=\sigma_i\tau_k \implies \tau_i=\tau_k$$ and $$\sigma_j\tau_i=\sigma_k\tau_i \implies \sigma_i=\sigma_k$$. Isn't it imply that all the $\sigma_i\tau_j$ unique?I really don't understand the thing in the red box. what are they trying to convince. Can you please explain?
Fix some $\mu \in S_{k+l}$. They are trying to explain that there are $k!$ ways to write $\mu$ as $\sigma \tau$, where $\sigma \in S_{k+l}, \tau \in S_k$ where $S_k$ is embedded in $S_{k+l}$ by fixing the last $l$ entries. (So in the sum, this one $\mu$ appears $k!$ times).
The reason for why there are $k!$ ways to write $\mu$ as $\sigma\tau$ is that there are $k!$ choices of $\tau$ since the size of $S_k$ is $k!$, and once you pick on $\tau \in S_k$, there is exactly one $\sigma \in S_{k+l}$ so that $\mu=\sigma \tau$. Namely, $\sigma$ has to be $\mu\tau^{-1}$