For each finite set $X$, the structure of the free module $F(X)$ is entirely determined by the cardinality of $X$. Let's consider some specific examples.

The free module on the empty set

What is the free module on the empty set? According to our construction the set of elements of the $R$-module $F(\mathrm{\beta \x88\x85})$ consists of all formal finite $R$-linear combinations of elements $\mathrm{\beta \x88\x85}$. But what the heck is the set of combinations of nothing, you might ask? Let's look at the universal property $F(\mathrm{\beta \x88\x85})$ should enjoy, namely that there is a natural bijection

The empty set is the initial object in $\mathbf{\text{Set}}$, so there is a unique set map (the empty map) from it to any other set. In other words, the set ${\mathrm{Hom}}_{\mathbf{\text{Set}}}(\mathrm{\beta \x88\x85},U(M))$ is a singleton set. Our bijection above then implies ${\mathrm{Hom}}_{R}(F(\mathrm{\beta \x88\x85}),M)$ is a singleton set, for every $R$-module $M$. This exactly says that $F(\mathrm{\beta \x88\x85})$ is the initial object in the category $R\mathbf{\text{-Mod}}$, which we've already seen is the zero module. Thus, $F(\mathrm{\beta \x88\x85})=0$, the zero module.

This is why you sometimes see books/people declare (usually by fiat) that the "empty combination" is the zero element.

Summary

The free module on the empty set is the zero module.

The free module on a singleton set

Now let's consider a singleton set $X=\{x\}$. By our construction, the module $F(\{x\})$ consists of all $R$-multiples $rx$ with $r\beta \x88\x88R$. We therefore have a set bijection $F(\{x\})\beta \x89\x83R$, which you can quickly verify is actually an $R$-module isomorphism. We could have already predicted this, since we've seen that ${\mathrm{Hom}}_{R}(R,M)\beta \x89\x83M$ as an $R$-module (at least for commutative rings $R$), and the universal property for $F(\{x\})$ is that we have a natural set bijection

For any singleton set $X$, the free module on $X$ is isomorphic to $R$ as an $R$-module.

The free module on a finite set

Next let's consider an arbitrary finite set $X=\{{x}_{1},{x}_{2},\beta \x80\xa6,{x}_{n}\}$. By our construction, the module $F(X)$ consists of all formal $R$-linear combinations of the form ${r}_{1}{x}_{1}+{r}_{2}{x}_{2}+\beta \x8b\u2015+{r}_{n}{x}_{n}$, which his evidently isomorphic to the $R$-module $R\beta \x8a\x95R\beta \x8a\x95\beta \x8b\u2015\beta \x8a\x95R$ (the direct sum of $R$ with itself $n$ times). This $R$-module is usually denoted ${R}^{n}$.

Is this bad notation?

The notation ${R}^{n}$ is usually shorthand for the direct product $R\Gamma \x97R\Gamma \x97\beta \x8b\u2015\Gamma \x97R$, not the direct sum. Fortunately, for finite families in the category $R\mathbf{\beta \x88\x92}\mathbf{Mod}$ the direct product and direct sum are isomorphic $R$-modules.