How to partition a set in order to minimize the number of the elements and their interactions?
Consider the following two variants of subset sum problem:
Variant 1

Given a set of $2n$ positive integer elements and a positive integer target $W$, is there a subset with size $n$, whose sum is $W$?

Variant 2

Given a set of $n$ positive integer elements and a positive integer target $W$, where $(\text{the maximum element})&lt; 2\times(\text{the minimum element})$, is there a subset whose sum is $W$?

Variant 1 is NP-hard by a reduction from one-in-three 3-SAT (almost the same reduction from 3SAT to the normal subset sum problem, except that we do not need the $s$ and $t$ values, and the target is always $111\ldots1$).
Variant 2 is also NP-hard by a reduction from Variant 1. Given an instance of Variant 1, we add a large enough integer $M$ to each element so that the new elements satisfy the constraint in Variant 2. Now the answer to Variant 1 is &quot;yes&quot; if and only if there is a subset of the new elements whose sum is $W+nM$ (since $M$ is large enough, the subset must have size $n$).
Now we show a reduction from Variant 2 to (the decision version of) your problem, so your problem is NP-hard.
Given an instance of Variant 2, we construct $S_1=S_2$ as the same set as the given instance. Let $s$ denote the sum of all elements. We set $R_1=W$ and $R_2=s-W$. The gain of each element is the same as itself.
Now if there is a subset $O\subseteq S_1=S_2$ whose sum is $W$, then we can choose $O_1=O$ and $O_2=S_2\backslash O$, meaning there is a solution to your problem with $|O_1|+|O_2|+|O_1\cap O_2|\le n$.
On the other hand, if there is a solution to your problem with $|O_1|+|O_2|+|O_1\cap O_2|\le n$, then $2|O_1\cap O_2|\le n-|O_1\cup O_2|$. Let $\min$ and $\max$ be the minimum and maximum elements respectively. We have
$$\sum_{e\in O_1\cap O_2}e\le |O_1\cap O_2|\max\le 2|O_1\cap O_2|\min\le(n-|O_1\cup O_2|)\min\le\sum_{e\notin O_1\cup O_2}e.$$
This means
$$
\sum_{e\notin O_1}e=\sum_{e\in O_2}e-\sum_{e\in O_1\cap O_2}e+\sum_{e\notin O_1\cup O_2}e\ge \sum_{e\in O_2}e\ge s-W.
$$
Note $\sum_{e\in O_1}e\ge W$ and $\sum_{e\notin O_1}e+\sum_{e\in O_1}e=s$, we have $\sum_{e\in O_1}e=W$ and $\sum_{e\notin O_1}e =s-W$, which means there is a subset $O\subseteq S_1=S_2$ whose sum is $W$.