I have a ring $R$ and a set $S$, I want to generate a free module with coefficients from $R$, such that elements are linear expressions of the form $$r_1 s_1 + r_2 s_2 + ... + r_n s_n,$$ where $r_i \in R$ and $s_i \in S$. Can I just say the free module of $R$ generated by the set $S$?
What I am confused about is that in the definition of a $R$-module, $S$ should have a group structure $(S,+)$. What if I do not have that, is it still well defined or do I get something that is not an $R$-module?
On
What we call the free $R$-module generated by the set $S$ is defined exactly the way you describe it. However, be careful: It is not $S$ that ends up being the module, so $(S, +)$ doesn't make sense. There is (a priori) no addition on $S$, and we don't define an addition on $S$. The underlying set of this free module, where we define addition and scalar multiplication, is the set $$ \{r_1s_1 + r_2s_2 + \cdots + r_ns_n\mid r_i\in R, s_i\in S\} $$ where we don't worry too much about what the symbol "$+$" really means, and just let it be implicit that it behaves the way we want addition to behave, so that for instance the element $r_1s_1 + r_2s_2$ is the same as the element $r_2s_2 + r_1s_1$. These are called formal linear combinations.
We define addition the straight-forward way, with for instance $$ (r_1s_1 + r_2s_2) + (r_3s_2 + r_4s_3) = r_1s_1 + (r_2+r_3)s_2 + r_4s_3 $$ (don't be confused; there are three different "$+$" symbols in there, meaning different things). Scalar multiplication is defined similarily: $$ r_1\cdot (r_2s_2 + r_3s_3) = (r_1r_2)s_2 + (r_1r_3)s_3 $$ For instance, the module $R\times R$ may be interpreted as the free $R$-module generated by the set $\{(1,0), (0,1)\}$.
On
The free module generated by $S$ is the $R$-module $$R^{(S)}=\bigl\{f\colon S\to R\mid f(s)=0,\;\text{ but for a finite number of }s\bigr\},$$ in other words, it is the set of maps from $S$ to $R$ with finite support.
A basis of this $R$-module is the set of maps $\mathbf 1_s$ defined by $$\mathbf 1_s(t)=\begin{cases}0&\text{if }t\ne s,\\1&\text{if }t=s.\end{cases}$$ Usually, these maps are denoted $\;\mathbf 1_s=[s]$, so that a generic element of $R^{(S)}$ be denoted $$\lambda_{1}[s_{1}]+\dots+ \lambda_{n}[s_{n}]$$ for some $n$.
If $S$ is a finite set, and $|S|=k$, then $R^{(S)}$ is (isomorphic to) $R^k$.
Note that the ring $R[X]$ of polynomials with coefficients in $R$ is defined as the $R$-module $R^{(\mathbf N)}$, on which a multiplicative structure is defined, so that it becomes an $R$-algebra.
Similarly, the ring of polynomials in $r$ indeterminates is the $R$-module $\;R^{(\mathbf N^r)}$ with a multiplicative structure.
It is still well defined. The confusion comes from the fact that $S$ is not the module itself, but is only an index set or set of generators for the free module. We define $(s_1) + (s_2)$ to be the symbol $(s_1+s_2)$. This means that for addition, you treat the elements $s_1 \ldots s_n$ like basis vectors. So, for example, $$(r_1 s_1 + r_2 s_2) + (r_3 s_2 + r_4s_3) = r_1 s_2 + (r_2+r_3)s_2 + r_3s_4.$$
An easy way to think about it is to think of your free module as being given by the set $R^S$ of functions $S \to R$. Then addition of functions and multiplication of functions by scalars (in $R$) give you the $R$-module structure.