Here, all rings are unital and $R$-modules are in fact left $R$-modules.
Let $M_1$ be a $R_1$-module and $M_2$ be a $R_2$-module. Then, we can view $M_1\oplus M_2$ as a ($R_1\oplus R_2$)-module, under the operation $\mathcal{o}((r_1,r_2).(m_1,m_2)):=(r_1m_1,r_2m_2)$.
My question is : Given any ($R_1\oplus R_2$)-module $M$, is $M$ isomorphic (as $(R_1\oplus R_2)$-module) to some $M'=M_1\oplus M_2$, where $M_i$ is a $R_i$-module and $M'$ is a $(R_1\oplus R_2)$-module under $o$ ?
Yes.
In this post, I will write $R_1\times R_2$ instead of $R_1\oplus R_2$; I hope you will not mind.
A common way to state the property you want (and slightly more) is the following: the module categories $\operatorname{Mod}(R_1 \times R_2)$ and $(\operatorname{Mod}R_1) \times (\operatorname{Mod}R_2)$ are equivalent.
However, we can prove the statement in your post directly. Let $M$ be an $R_1\times R_2$-module. Consider the subset $$ (1_{R_1}, 0)M := \{(1_{R_1}, 0)m \ | \ m\in M\} $$ of $M$. It is easily seen that it is an $R_1$-module for the product $$ r\cdot m = (r,0)m. $$ Similarly, $(0,1_{R_2})M$ is naturally an $R_2$-module.
Then $M$ is isomorphic as an $R_1\times R_2$-module to $(1_{R_1}, 0)M \oplus (0,1_{R_2})M$, with the product described by the $o$ in your post.
