For permutation $\sigma$ let $\sigma T(X_1, \ldots, X_n) = T(X_{\sigma(1)},\ldots, X_{\sigma(n)})$, why then $\tau(\sigma T) = (\tau\sigma) T$? I've encountered this first in Lang's Algebra (believe me, I've mastered major parts of that book), but the first notation is actually from Lee (Introduction to Smooth Manifolds, Chapter 11. Tensors) where $T$ is a covariant tensor and the goal is to symmetrize it, so we define
$$S = \frac{1}{k!}\sum_{\sigma\in S_k}\sigma T$$
and it is easy to see with the fact above that this tensor is symmetric.
Anyway, back to Lang notation (page 30., after symmetric groups and some examples), let
$$\pi(\sigma)f(x_1, \ldots, x_n) = f(x_{\sigma(1)}, \ldots, x_{\sigma(n)})$$
we calculate
$$\pi(\sigma)\pi(\tau)f(x_1, \ldots, x_n) = (\pi(\tau)f)(x_{\sigma(1)}, \ldots, x_{\sigma(n)}) = f(x_{\sigma\tau(1)},\ldots, x_{\sigma\tau(n)}) = \pi(\sigma\tau)f(x_1,\ldots, x_n)$$
First and last equality are from the definition, but I just cannot grasp my head around second equality, I thought it needs to be reversed, $\tau\sigma$. It frustrates me that I cannot understand this trivial elementary calculation while I easily understand some harder concepts.
Can you please explain it like I'm 5 years old?

Let $(y_1,\dots,y_n):=(x_{\sigma(1)},\dots,x_{\sigma(n)}).$
$(\pi(\tau)f)(x_{\sigma(1)}, \ldots, x_{\sigma(n)})=(\pi(\tau)f)(y_1,\dots,y_n)=f(y_{\tau(1)},\dots,y_{\tau(n)}).$
Since  $y_k=x_{\sigma(k)}$ for all $k,$ what are the $y_{\tau(j)}$'s equal to, dear child?
