Let $R$ be a ring and ruppose $F$ is a set of all functions from $S$ to $R$. Then, every element $u\in S$ can be viewed as a function from $S$ to $R$: $$u(t)=\begin{cases} 1 \hbox{ if } t=u\\ 0 \hbox{ if } t\neq u \end{cases}$$ So $S$ is a subset of $F$.
Note that $F$ is an abelian group with $$[f+g](t):=f(t)+g(t)$$ for all $f,g\in F$, and $t\in S$.
I'm trying to prove that $F$ is a right $R$-module with scaling operation $$[f\cdot r](t):=f(t)r$$ for all $f\in F$, $r\in R$ and $t\in S$, but I'm having trouble how to show this.
To show the required claim we have to check that for any $f,g\in F$, $r,s\in R$, and $t\in S$ we have
1)) $[(f+g)\cdot r](t)=([f\cdot r]+ [g\cdot r])(t)$
2)) $[f\cdot (r+s)](t)=([f\cdot r]+ [f\cdot s])(t)$.
For (1) we have
$[(f+g)\cdot r](t)=([f+g](t))r=(f(t)+g(t))r=f(t)r+g(t)r=[f\cdot r](t)+[g\cdot r](t)=([f\cdot r]+ [g\cdot r])(t)$
For (2) we have
$[f\cdot (r+s)](t)=f(t)(r+s)=f(t)r+f(t)s=[f\cdot r](t)+ [f\cdot s](t)=([f\cdot r]+ [f\cdot s])(t)$.