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?
 A: I think the whole thing revolves around understanding the order in which permutations and evaluations by $f$ occur.

*

*In $g:=\phi(\tau)f$, first we permute the order of the elements of  $n$-tupple argument according to $\tau$, and then  evaluate by $f$.


*In $\pi(\sigma)\pi(\tau)f$ we apply $\pi(\sigma)$ to the function $\phi(\tau)f$, which can be written as $\pi(\sigma)\big(\pi(\tau)f\big)$.  That is, first the permute the elements of the $n$-tuple argument according to $\tau$, and then evaluate by the function $\pi(\tau)f$.
Combining 1 and 2 yields:
$$(\pi(\sigma)\pi(\tau)f )(x_1,\ldots,x_n)=\big(\pi(\tau)f\big)(x_{\sigma(1)},\ldots,x_{\sigma(n)})=g(y_1,\ldots,y_n)$$
Let $g:=\pi(\tau)f$, and let $y_j=x_{\sigma(j)}$, $1\leq j\leq n$. Notice that the subscript $j$ in $y$ turns into a subscript $\sigma(j)$ in $x$.
Then
\begin{align}
g(y_1,\ldots,y_n)&=\big(\pi(\tau)f\big)(y_1,\ldots,y_n)\\
&=f(y_{\tau(1)},\ldots, y_{\tau(n)})\\
&=f(x_{\sigma(\tau(1))}, \ldots, y_{\sigma(\tau(n))})\\
&=\big(\pi(\sigma\circ \tau)f\big)(x_1,\ldots, x_n)
\end{align}
That is
$$\pi(\sigma)\pi(\tau)f=\pi(\sigma\circ\tau)f$$
A: I'll try to explain how I think about this; it won't be suitable for a 5-year-old, but it might be suitable for you or for other readers.
As a preliminary step, consider a set $A$ on which a group $G$ acts (from the left) and another set $B$ (with no additional structure). Then $G$ also acts on the set $B^A$ of functions from $B$ to $A$ by composition. If we write this action on the left, as $\pi f=f\circ\pi$ (for $\pi\in G$ and $f\in B^A$), then we get just the sort of order-reversal that you're worried about:
$$
\pi(\sigma f)=\pi(f\circ\sigma)=f\circ\sigma\circ\pi 
=(\sigma\pi)f.
$$
In your situation, three sets are involved: The set $I=\{1,2,\dots,n\}$ of indices, the set $V$ of possible values of the $x$'s in the question, and the range $R$ of the function $f$. The symmetric group $S_n$ acts on $I$. It therefore also acts, as in the preceding paragraph, on the set $V^I$ of $n$-tuples $(x_1,x_2,\dots,x_n)$ of values, because such an $n$-tuple is exactly a function from $I$ into $V$. Note that this $V^I$ is the domain of your $f$. The action of $S_n$ on $V^I$ induces, again as in the preceding paragraph, an action on $R^{V^I}$, the space in which your $f$ and its shuffled versions live.
As in the preliminary paragraph, passing from an action on $A$ to one on $B^A$ reverses the order in the multiplication. (One could say that a left action on $A$ induces a right action on $B^A$.) In your situation, there are two such reversals, first in passing from $I$ to $V^I$ and second in passing from $V^I$ to $R^{V^I}$. And of course reversing the order of multiplication twice restores the original order. (A left action on $I$ induces a left action on $R^{V^I}$.)
A: 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?
