Definition: A module $M$ over a commutative ring $R$ is said to be finitely generated by $m_1,\dots,m_n \in M$ if for any $m \in M$, there exist $r_1,\dots, r_n \in R$ such that $m = \sum_{i=1}^n r_i m_i$.
My question is that does the above definition imply $$M = \sum_{i=1}^n R m_i,$$ where $R m_i:= \{m \mid m = r m_i, r\in R\}?$
The definition in the first paragraph only states the exists of $r_i$ for a given $m$, but it seems not to exclude the situation where there may exist $r_1,\dots,r_n \in R$ such that $\sum_{i=1}^n r m_i$ is not in $M$. Is there a proof needed for the property under question? Or is the property under question self-evident from the definition?
If the property does not hold, do I need to impose additional conditions on the $R$-module $M$? For example, by restricting it to be a free module or by restricting $R$ to be a principal ideal domain?
For a left $R$ module $M$ generated by $m_1,\ldots, m_n$, it is true that $\sum_{i=1}^n Rm_i=\left\{\sum_{i=1}^n r_i m_i\mid r_i\in R\right\}$, if that's what you're asking. As far as I know, that's the definition everyone is given.
In general it won't make sense to put elements of $r$ on the right of elements of $m$ since no such operation was defined. But if, say, $R$ is commutative, then you could consider it as a left and right module and move the ring elements to either side.
Typically though people are going to pick one side and stick with it.