Show that group action is homomorphism to Symmetric group

Although your notation is correct, I really dislike it -- I think it obfuscates this very clear result. If we just denote the map $G \times X \to X$ by juxtaposition, the group action axioms become \begin{align*} 1 x &= x\\ g_1(g_2 x) &= (g_1 g_2) x \end{align*} for all $x \in X$ and $g_1, g_2 \in G$. For each $g \in G$, define \begin{align*} \sigma_g : X &\to X\\ x &\mapsto gx \end{align*} which is a bijection $X \to X$, i.e., an element of $\text{Sym(X)}$. Define the map \begin{align*} \varphi : G &\to \text{Sym}(X)\\ g &\mapsto \sigma_g \, . \end{align*} Given $x \in X$, the second group action axiom then gives \begin{align*} \varphi(g_1 g_2)(x) &= \sigma_{g_1 g_2}(x) = (g_1 g_2)x = g_1(g_2 x) = g_1(\sigma_{g_2} (x)) = \sigma_{g_1}(\sigma_{g_2}(x))\\ &= (\sigma_{g_1} \circ \sigma_{g_2})(x) = (\varphi(g_1) \circ \varphi(g_2))(x) \, . \end{align*} Thus $\varphi(g_1 g_2) = \varphi(g_1) \circ \varphi(g_2)$, so $\varphi$ is a homomorphism.

I'm denoting both the group action and the binary operation of the group by juxtaposition, so there is some chance of confusion. But I find it suggestive rather than confusing. The action has to be "compatible" with the group binary operation.

Note that * is the group operation on $Sym(X)$. So * represents composition of functions. Replacing the notation for $A(g,.) $ by $A_g$, we see that $A_{g+f}(x)= A(g+f,x)= A(g,A(f,x))=A(g,A_f(x))=A_g*A_f(x)$ which is what we want. Note again that * represents composition of functions.