Let $A$ be a commutative ring with unity, and $S_1,S_2$ multiplicative subsets. Let $\lambda_i:A\rightarrow A_{S_i}$, $x\mapsto x/1$ be the natural ring homomorphism. For an ideal $\mathfrak{a}$ in $A$, let $\mathfrak{a}^e$ denote the extension of $\mathfrak{a}$ under localization; it is ideal generated by image, under natural localizing map, of $\mathfrak{a}$. Similarly, for an ideal $\mathfrak{b}$ in $A_{S}$ let $\mathfrak{b}^c$ denote the contraction of ideal $\mathfrak{b}$, i.e. the inverse image under natural localizing map.(The lextension and contraction are w.r.t. a fixed localization). Let $S(\mathfrak{a}):=(\mathfrak{a}^e)^c$.
It is well known that $$(*)\,\,\,\,(\mathfrak{a}^e)^c=\cup_{s\in S} (\mathfrak{a}:s).$$
Exercise: If $S_1,S_2$ are multiplicatively closed subsets of $A$ then $S_1(S_2(\mathfrak{a}))=(S_1S_2)(\mathfrak{a}).$
My answer is simply using above fact $(*)$: $$S_1(S_2(\mathfrak{a}))=S_1(\cup_{s_2\in S_2}(\mathfrak{a}:s_2))=\cup_{s_1\in S_1}\cup_{s_2\in S_2}((\mathfrak{a}:s_2):s_1)=\cup_{s_1s_2\in S_1S_2}(\mathfrak{a}:s_1s_2).$$
Is this correct?