Let $R$ be a commutative ring with identity $1$. The definition of a (unital) R-module is given as an abelian group $M$ together with an operation $R \times M \to M$, usually just written as $rv$ when $r \in R$ and $v \in M$. The scaling operation satisfies the following conditions: \begin{align} 1v = v ~~~\text{ for all }~~~~ v \in M \\ (rs)v = r(sv)~~~\text{for all }r,s \in R \text{ and all }~v \in M \\ (r+s)v = rv + sv \text{ for all } r,s \in R \text{ and all }~ v\in M \\ r(v+w) = rv + rw~~\text{ for all }r \in R ~~\text{ and }v,w \in M \end{align} It is then stated that "If $M$ is a R-module, so is the set of finite formal linear combinations $L(M)$ of elements of $M$, where in $L(M)$ you are not allowed to simplify, so for instance, $r(v)$ and $1(rv)$ are considered distinct elements if $r \neq 1$."
Is the idea here to endow $L(M)$ with the same operation as $M$ making $L(M)$ an abelian group (as it is currently just a set) and to define the scaling operation $R \times L(M) \to L(M)$ in the same way as we do $R \times M \to M$?
You won't give $L(M)$ the same addition structure as $M$. An arbitrary element of $L(M)$ has the form $\ell = \sum_{i = 1}^n r_i [m_i]$, where $m_i\in M$, or if you prefer $\ell = \sum_{m\in M} r_m [m]$, but writing in this second way means you must demand that all but finitely many $r_m$ are equal to $0$. Then to add two elements $ \ell = \sum_{m\in M} r_m [m]$ and $\ell' = \sum_{m\in M} r_m' [m]$, you simply add the coefficients of each $[m]$: no performing the addition in $M$ at all! $$ \ell + \ell' = \sum_{m\in M} r_m [m]+\sum_{m\in M} r_m' [m] := \sum_{m\in M} (r_m + r'_m) [m] $$ (one should check that all but finitely many of the $r_m + r'_m$ are equal to $0$ to ensure that the sum is a well-defined element of $L(M)$. And now, the action of $R$ on $L(M)$ is just scaling on the coefficients.
This construction does not depend on $M$ having a $R$-module structure; you can form $L(S)$ for any set $S$ in the same way, and this will have the same addition operation. In fact, if $\#S = \#M$, then $L(M)\cong L(S)$ (see if you can find the isomorphism).