I want to give an example of a module over a Boolean ring $R$ (in particular, the power set of a set $X$ under symmetric difference and set intersection).
I thought of defining a map $(Y, m) \mapsto |Y|m$ where $m$ is an element of an abelian group $M$. But that violates the module axiom: $(r+s)m=rm+sm$ since $|Y \Delta Z | \leq |Y|+|Z|$, so we can't guarantee equality.
Please note, I want $M \not\subseteq R$ since of course, $R$ and any of its ideals are $R$-modules.
For any ring $R$, $M = R^n$ is an $R$-module, and $M\not\subseteq R$ when $n>1$. You can get lots of other examples (all the finitely generated $R$-modules) by looking at quotients of $R^n$ by arbitrary $R$-submodules. There's nothing special about Boolean rings here.
In the Boolean case, there are some restrictions on what an $R$-module can look like. For example, if $R$ is a Boolean ring, and $M$ is an $R$-module, then we must have $x + x = 0$ for all $x\in M$. This is because $x + x = 1x + 1x = (1 + 1)x = 0x = 0$.
On the other hand, given any abelian group $M$ with $x+x = 0$ for all $x\in M$, we can give $M$ the structure of an $R$-module for any nontrivial Boolean ring $R$. First, we can view $M$ as a vector space over $\mathbb{F}_2$ (by $1x = x$ and $0x = 0$). And if we pick a maximal ideal $\mathfrak{m}\subseteq R$, we get the quotient map $\pi\colon R\to R/\mathfrak{m}\cong \mathbb{F}_2$, and we can define the $R$-module structure on $M$ by $r\cdot x = \pi(r)x$.
Concretely, if $R = \mathcal{P}(X)$ is a powerset ring, a maximal ideal on $R$ is the same thing as an ultrafilter $U$ on $X$, and for $Y\subseteq X$, $\pi(Y) = 1$ if $Y\in U$ and $\pi(Y) = 0$ if $Y\notin U$. So the $R$-module structure is defined by $$Y\cdot x = \begin{cases} x & \text{if }Y\in U\\ 0 & \text{if }Y\notin U\end{cases}$$ The most concrete kind of ultrafilter is the principal kind: Fix $p\in X$ and take $U_p = \{Y\subseteq X\mid p\in Y\}$. If $X$ is finite, then these are the only ultrafilters. So with $U = U_p$, we get $$Y\cdot x = \begin{cases} x & \text{if }p\in Y\\ 0 & \text{if }p\notin Y.\end{cases}$$