Let $\operatorname{Sym}(x)$ be the symetric group and then $\operatorname{Sym}(2)=\{id,(1,2)\}$. Let $X=\{1,2,3\}$. (This might be wrong as i think this set must only contain $1$ and $2$ as elements.)
Define the function $f:\operatorname{Sym}(2)\times X\to X$, where $f(x,y)=x\circ y$ and $\circ$ is the composition of maps.
So therefore $f:\{(id,1),(id,2),(id,3),((1,2),1),((1,2),2),((1,2),3)\}\to\{1,2,3\}$.
The question now it is how can I apply an action on an element of the domain. So for example $f((1,2),2)=(1,2)\circ 2=$?
This has been taken from lecture notes as being an example written as: $S_n$ acts on $\{1, 2, . . . , n\}$ via application of each map.
If I understood you correctly, $\operatorname{Sym}(2)$ acts on $X$ via $f$ by defining $f: \operatorname{Sym}(2)\times X\to X$ by $(\alpha, x)\mapsto \alpha\circ x$; that is, $f(\alpha, x)=\alpha\circ x\stackrel{\operatorname{Def}}{=}\alpha(x)$, where $\alpha(x)$ is $\alpha$ evaluated at $x\in X$.
It follows that here $X$ can be any set with $\{1,2\}\subseteq X$.
Using your example, $f((12), 2)=(12)\circ 2\stackrel{\operatorname{Def}}{=}(12)(2)=1$.
NB: Here $\stackrel{\operatorname{Def}}{=}$ means "is, by definition, equal to".