The functor $a_X^{-1}$ is fully faithful (as $a_{X, *}a_X^{-1} = \text{id}$ by Lemma \ref{lemma-describe-pullback-pi-fppf}). Hence the forgetful functor $(\mathcal{G}, \mathcal{H}, \alpha) \mapsto \mathcal{H}$ identifies the category of triples with a full subcategory of $\Sh((\Sch/Y)_{fppf})$. Moreover, the functor $a_Y^{-1}$ is fully faithful, hence the functor in the lemma is fully faithful as well. \medskip\noindent Suppose that we have an \'etale covering $\{Y_i \to Y\}$. Let $f_i : X_i \to Y_i$ be the base change of $f$. Denote $f_{ij} = f_i \times f_j : X_i \times_X X_j  \to Y_i \times_Y Y_j$. Claim: if the lemma is true for $f_i$ and $f_{ij}$ for all $i, j$, then the lemma is true for $f$. To see this, note that the given \'etale covering determines an \'etale covering of the final object in each of the four sites $Y_\etale, X_\etale, (\Sch/Y)_{fppf}, (\Sch/X)_{fppf}$. Thus the category of sheaves is equivalent to the category of glueing data for this covering (Sites, Lemma \ref{sites-lemma-mapping-property-glue}) in each of the four cases. A huge commutative diagram of categories then finishes the proof of the claim. We omit the details. The claim shows that we may work \'etale locally on $Y$. \medskip\noindent Note that $\{X \to Y\}$ is an fppf covering. Working \'etale locally on $Y$, we may assume there exists a morphism $s : X' \to X$ such that the composition $f' = f \circ s : X' \to Y$ is surjective finite locally free, see More on Morphisms, Lemma \ref{more-morphisms-lemma-dominate-fppf-etale-locally}. Claim: if the lemma is true for $f'$, then it is true for $f$. Namely, given a triple $(\mathcal{G}, \mathcal{H}, \alpha)$ for $f$, we can pullback by $s$ to get a triple $(s_{small}^{-1}\mathcal{G}, \mathcal{H}, s_{big, fppf}^{-1}\alpha)$ for $f'$. A solution for this triple gives a sheaf $\mathcal{F}$ on $Y_\etale$ with $a_Y^{-1}\mathcal{F} = \mathcal{H}$. By the first paragraph of the proof this means the triple is in the essential image. This reduces us to the case described in the next paragraph. \medskip\noindent Assume $f$ is surjective finite locally free. Let $(\mathcal{G}, \mathcal{H}, \alpha)$ be a triple. In this case consider the triple $$ (\mathcal{G}_1, \mathcal{H}_1, \alpha_1) = (f_{small}^{-1}f_{small, *}\mathcal{G}, f_{big, fppf, *}f_{big, fppf}^{-1}\mathcal{H}, \alpha_1) $$ where $\alpha_1$ comes from the identifications \begin{align*} a_X^{-1}f_{small}^{-1}f_{small, *}\mathcal{G} & = f_{big, fppf}^{-1}a_Y^{-1}f_{small, *}\mathcal{G} \\ & = f_{big, fppf}^{-1}f_{big, fppf, *}a_X^{-1}\mathcal{G} \\ & \to f_{big, fppf}^{-1}f_{big, fppf, *}f_{big, fppf}^{-1}\mathcal{H} \end{align*} where the third equality is Lemma \ref{lemma-proper-push-pull-fppf-etale} and the arrow is given by $\alpha$. This triple is in the image of our functor because $\mathcal{F}_1 = f_{small, *}\mathcal{F}$ is a solution (to see this use Lemma \ref{lemma-proper-push-pull-fppf-etale} again; details omitted). There is a canonical map of triples $$ (\mathcal{G}, \mathcal{H}, \alpha) \to (\mathcal{G}_1, \mathcal{H}_1, \alpha_1) $$ which uses the unit $\text{id} \to f_{big, fppf, *}f_{big, fppf}^{-1}$ on the second entry (it is enough to prescribe morphisms on the second entry by the first paragraph of the proof). Since $\{f : X \to Y\}$ is an fppf covering the map $\mathcal{H} \to \mathcal{H}_1$ is injective (details omitted). Set $$ \mathcal{G}_2 = \mathcal{G}_1 \amalg_\mathcal{G} \mathcal{G}_1\quad \mathcal{H}_2 = \mathcal{H}_1 \amalg_\mathcal{H} \mathcal{H}_1 $$ and let $\alpha_2$ be the induced isomorphism (pullback functors are exact, so this makes sense). Then $\mathcal{H}$ is the equalizer of the two maps $\mathcal{H}_1 \to \mathcal{H}_2$. Repeating the arguments above for the triple $(\mathcal{G}_2, \mathcal{H}_2, \alpha_2)$ we find an injective morphism of triples $$ (\mathcal{G}_2, \mathcal{H}_2, \alpha_2) \to (\mathcal{G}_3, \mathcal{H}_3, \alpha_3) $$ such that this last triple is in the image of our functor. Say it corresponds to $\mathcal{F}_3$ in $\Sh(Y_\etale)$. By fully faithfulness we obtain two maps $\mathcal{F}_1 \to \mathcal{F}_3$ and we can let $\mathcal{F}$ be the equalizer of these two maps. By exactness of the pullback functors involved we find that $a_Y^{-1}\mathcal{F} = \mathcal{H}$ as desired.