Let a group $G$ acts on a vector space $V$ and let $f$ be a function on $V$. The action of an element $g \in G$ defined by the rule $g f(x)=f(g^{-1} x), \forall x \in V.$
A typical proof from a book has the form $$ g_1 g_2 f(x)=g_1 f(g_2^{-1} x)=f(g_2^{-1}g_1^{-1} x)=f((g_1 g_2)^{-1} x) $$
Question. Why here $g_1 f(g_2^{-1} x)=f(g_2^{-1}g_1^{-1} x)$? Seems have to be $g_1 f(g_2^{-1} x)=f(g_1^{-1}g_2^{-1} x)$.
Let me clarify the confusion. Elements $g \in G$ act on the functions on $V$. Thus, if we have
$$g_{1} \cdot f(g_{2}^{-1} x)$$
we need to identify what the the function is. Is the function here $f$? No, because the function is meant to take as input elements of $V$, but inside of $f$ we have a group element $g_{2}^{-1}$. The actual function here is
$$h = f \circ g_{2}^{-1}$$
since this takes as input $x \in V$. Thus
$$g_{1} \cdot h(x) = h(g_{1}^{-1}x) = f \circ g_{2}^{-1} \circ g_{1}^{-1} (x)$$