In this explanation, $L$ is an operator that could be the Laplacian. However, I don't know what is $f$ and $T$. It later suggests that $T$ is a rotation, but what is $f$ and why does the relation $L(f\circ T) = (Lf)\circ T$ looks like associativity, not commutativity?
Where it says “for any translation (rotation) on $\mathbb R^n$”, $T$ is that translation or rotation, so $T$ stands for translation and rotation operators. $f$ stands for a function.
That the equation looks more like associativity than like commutativity is due to the somewhat unhelpful notation that they're using. The operator $L$ is being applied from the left and without a symbol, and the translation / rotation operator $T$ is being applied from the right and with the symbol $\circ$. If we use the same notation for these operators, the equation becomes
$$ L(Tf)=T(Lf)\;, $$
and if we now define the product $AB$ of operators $A$ and $B$ via $(AB)f=A(Bf)$, then this becomes
$$ (LT)f=(TL)f $$
and thus
$$ LT=TL $$
(since $f$ is arbitrary, so the equality for all $f$ implies the equality of the operators).
Thus, that $L$ commutes with translations means that it doesn't matter whether you first apply the translation and then $L$ or the other way around.