Algorithm

We abbreviate $\Omega:=\Omega_{X,0}$, $\mathcal{H}'':=\mathcal{H}''_0$, $\mathcal{G}:=\mathcal{G}_0$, $s:=\partial^{-1}_t$, and we consider the rings ${\mathbf{C}\{t\}}$ and ${\mathbf{C}\{\!\{s\}\!\}}$. Then $[s^{-2}t,s]=1$ and, hence, $t=s^2\partial_s$.