The following statement seems true, but I don't know a proof or a reference for it (and I would like one).

Let $\Gamma< \operatorname{PSL}(2,\mathbb R)$ be a nonuniform lattice with one cusp. We may conjugate $\Gamma$ so that an element $ \begin{pmatrix} 1 & s\\ 0 & 1 \end{pmatrix} $ generates the cusp of $\Gamma$. Let $\Gamma'$ be such that $\Gamma<\Gamma'$ be an index $2$ subgroup. Then, the element $ \begin{pmatrix} 1 & s/2\\ 0 & 1 \end{pmatrix} $ generates the cusp of $\Gamma'$.

I imagine a similar statement would be true if both $\Gamma$ and $\Gamma'$ had two cusps.