Suppose that we have a finite group $G$ with three subgroups $A,B,C$. I am interested in relating the action of $A \cap B \cap C$ on $B \cap C$ with the action of $A \cap B$ on $B$ (both via right multiplication). In particular, I am interested in relating transversals (a collection of representatives for the orbit space) for these actions.
I'm not sure exactly what we might be able to say in this situation. I think the following may be true:
There exists a transversal $T \subseteq B$ for the coset space $B / (A \cap B)$ such that $T \cap C$ is a transversal for $(B \cap C) / (A \cap B \cap C)$.
and also perhaps the stronger statement
Let $T' \subseteq B \cap C$ be a transversal for the coset space $(B \cap C) / (A \cap B \cap C)$. Then there exists a transversal $T' \subseteq T \subseteq B$ for the coset space $B / (A \cap B)$ such that $T \cap C = T'$.
Now I'm not sure if we could go further and say something like:
If $T \subseteq B$ is a transversal for the coset space $B / (A \cap B)$, then $T \cap C$ is a transversal for $(B \cap C) / (A \cap B \cap C)$.
or perhaps
If $T \subseteq B$ is a transversal for the coset space $B / (A \cap B)$, then $T \cap C$ is a transversal for $(B \cap C) / (A \cap B \cap C)$, and every such transversal is obtained in this way.
If not, is there something we can say generally about this set up? Perhaps something like:
We assume that $A,C\leq B$. Let $B$ be three copies of $C_2$, generated by $x,y,z$. Let $A=\langle x,y\rangle$. Then $T=\{1,z\}$ is a transversal to $A$ in $B$. Now let $C=\langle y,z\rangle$. Then $A\cap C=\langle y\rangle$, and $T=T\cap C$ is indeed a transversal to $A\cap C$ in $C$.
But $T=\{1,xz\}$ is also a transversal to $A$ in $B$. And it has a different intersection with $C$, namely $\{1\}$.
So, in conclusion, your statement does not hold, that $T\cap C$ is a transversal to $A\cap C$ in $B$. Thus extend $T'$ to $T$ in any way.
Edit: You can however always choose such a transversal. Simply choose a transversal $T$ such that $T\cap C$ is a transversal to $A\cap B\cap C$ in $B\cap C$. Choose a transversal $T'$ to $A\cap B\cap C$ in $B\cap C$. If $x$ and $y$ are different coset representatives, then $xy^{-1}\not\in A\cap B\cap C$. If $xy^{-1}\in A\cap B$ though, then $xy^{-1}\in A\cap B\cap C$ (as $x,y\in C$), a contradiction. Thus they label different cosets of $A\cap B$ in $B$.