The category of compactifications of $X$ over $Y$ is defined in More on Flatness, Section \ref{flat-section-compactify}. By More on Flatness, Theorem \ref{flat-theorem-nagata} and Lemma \ref{flat-lemma-compactifyable} it is nonempty. To every choice of a compactification $$ j : X \to \overline{X},\quad \overline{f} : \overline{X} \to Y $$ the construction above associates the functor $j^* \circ \overline{a} : D_\QCoh^+(\mathcal{O}_Y) \to D_\QCoh^+(\mathcal{O}_X)$ where $\overline{a}$ is the right adjoint of $R\overline{f}_*$ constructed in Lemma \ref{lemma-twisted-inverse-image}. \medskip\noindent Suppose given a morphism $g : \overline{X}_1 \to \overline{X}_2$ between compactifications $j_i : X \to \overline{X}_i$ over $Y$ such that $g^{-1}(j_2(X)) = j_1(X)$\footnote{This may fail with our definition of compactification. See More on Flatness, Section \ref{flat-section-compactify}.}. Let $\overline{c}$ be the right adjoint of Lemma \ref{lemma-twisted-inverse-image} for $g$. Then $\overline{c} \circ \overline{a}_2 = \overline{a}_1$ because these functors are adjoint to $R\overline{f}_{2, *} \circ Rg_* = R(\overline{f}_2 \circ g)_*$. By (\ref{equation-sheafy}) we have a canonical transformation $$ j_1^* \circ \overline{c} \longrightarrow j_2^* $$ of functors $D^+_\QCoh(\mathcal{O}_{\overline{X}_2}) \to D^+_\QCoh(\mathcal{O}_X)$ which is an isomorphism by Lemma \ref{lemma-proper-noetherian}. The composition $$ j_1^* \circ \overline{a}_1 \longrightarrow j_1^* \circ \overline{c} \circ \overline{a}_2 \longrightarrow j_2^* \circ \overline{a}_2 $$ is an isomorphism of functors which we will denote by $\alpha_g$. \medskip\noindent Consider two compactifications $j_i : X \to \overline{X}_i$, $i = 1, 2$ of $X$ over $Y$. By More on Flatness, Lemma \ref{flat-lemma-compactifications-cofiltered} part (b) we can find a compactification $j : X \to \overline{X}$ with dense image and morphisms $g_i : \overline{X} \to \overline{X}_i$ of compactififcatins. By More on Flatness, Lemma \ref{flat-lemma-compactifications-cofiltered} part (c) we have $g_i^{-1}(j_i(X)) = j(X)$. Hence we get isomorpisms $$ \alpha_{g_i} : j^* \circ \overline{a} \longrightarrow j_i^* \circ \overline{a}_i $$ by the previous paragraph. We obtain an isomorphism $$ \alpha_{g_2} \circ \alpha_{g_1}^{-1} : j_1^* \circ \overline{a}_1 \to j_2^* \circ \overline{a}_2 $$ To finish the proof we have to show that these isomorphisms are well defined. We claim it suffices to show the composition of isomorphisms constructed in the previous paragraph is another (for a precise statement see the next paragraph). We suggest the reader check this is true on a napkin, but we will also completely spell it out in the rest of this paragraph. Namely, consider a second choice of a compactification $j' : X \to \overline{X}'$ with dense image and morphisms of compactifications $g'_i : \overline{X}' \to \overline{X}_i$. By More on Flatness, Lemma \ref{flat-lemma-compactifications-cofiltered} we can find a compactification $j'' : X \to \overline{X}''$ with dense image and morphisms of compactifications $h : \overline{X}'' \to \overline{X}$ and $h' : \overline{X}'' \to \overline{X}'$. We may even assume $g_1 \circ h = g'_1 \circ h'$ and $g_2 \circ h = g'_2 \circ h'$. The result of the next paragraph gives $$ \alpha_{g_i} \circ \alpha_h = \alpha_{g_i \circ h} = \alpha_{g'_i \circ h'} = \alpha_{g'_i} \circ \alpha_{h'} $$ for $i = 1, 2$. Since these are all isomorphisms of functors we conclude that $\alpha_{g_2} \circ \alpha_{g_1}^{-1} = \alpha_{g'_2} \circ \alpha_{g'_1}^{-1}$ as desired. \medskip\noindent Suppose given compactifications $j_i : X \to \overline{X}_i$ for $i = 1, 2, 3$.  Suppose given morphisms $g : \overline{X}_1 \to \overline{X}_2$ and $h : \overline{X}_2 \to \overline{X}_3$ of compactifications such that $g^{-1}(j_2(X)) = j_1(X)$ and $h^{-1}(j_2(X)) = j_3(X)$. Let $\overline{a}_i$ be as above. The claim above means that $$ \alpha_g \circ \alpha_h = \alpha_{g \circ h} : j_1^* \circ \overline{a}_1 \to j_3^* \circ \overline{a}_3 $$ Let $\overline{c}$, resp.\ $\overline{d}$ be the right adjoint of Lemma \ref{lemma-twisted-inverse-image} for $g$, resp.\ $h$. Then $\overline{c} \circ \overline{a}_2 = \overline{a}_1$ and $\overline{d} \circ \overline{a}_3 = \overline{a}_2$ and there are canonical transformations $$ j_1^* \circ \overline{c} \longrightarrow j_2^* \quad\text{and}\quad j_2^* \circ \overline{d} \longrightarrow j_3^* $$ of functors $D^+_\QCoh(\mathcal{O}_{\overline{X}_2}) \to D^+_\QCoh(\mathcal{O}_X)$ and $D^+_\QCoh(\mathcal{O}_{\overline{X}_3}) \to D^+_\QCoh(\mathcal{O}_X)$ for the same reasons as above. Denote $\overline{e}$ the right adjoint of Lemma \ref{lemma-twisted-inverse-image} for $h \circ g$. There is a canonical transformation $$ j_1^* \circ \overline{e} \longrightarrow j_3^* $$ of functors $D^+_\QCoh(\mathcal{O}_{\overline{X}_3}) \to D^+_\QCoh(\mathcal{O}_X)$ given by (\ref{equation-sheafy}). Spelling things out we have to show that the composition $$ \alpha_h \circ \alpha_g : j_1^* \circ \overline{a}_1 \to j_1^* \circ \overline{c} \circ \overline{a}_2 \to j_2^* \circ \overline{a}_2 \to j_2^* \circ \overline{d} \circ \overline{a}_3 \to j_3^* \circ \overline{a}_3 $$ is the same as the composition $$ \alpha_{h \circ g} : j_1^* \circ \overline{a}_1 \to j_1^* \circ \overline{e} \circ \overline{a}_3 \to j_3^* \circ \overline{a}_3 $$ We split this into two parts. The first is to show that the diagram $$ \xymatrix{ \overline{a}_1 \ar[r] \ar[d] & \overline{c} \circ \overline{a}_2 \ar[d] \\ \overline{e} \circ \overline{a}_3 \ar[r] & \overline{c} \circ \overline{d} \circ \overline{a}_3 } $$ commutes where the lower horizontal arrow comes from the identification $\overline{e} = \overline{c} \circ \overline{d}$. This is true because the corresponding diagram of total direct image functors $$ \xymatrix{ R\overline{f}_{1, *} \ar[r] \ar[d] & Rg_* \circ R\overline{f}_{2, *} \ar[d] \\ R(h \circ g)_* \circ R\overline{f}_{3, *} \ar[r] & Rg_* \circ Rh_* \circ R\overline{f}_{3, *} } $$ is commutative (insert future reference here). The second part is to show that the composition $$ j_1^* \circ \overline{c} \circ \overline{d} \to j_2^* \circ \overline{d} \to j_3^* $$ is equal to the map $$ j_1^* \circ \overline{e} \to j_3^* $$ via the identification $\overline{e} = \overline{c} \circ \overline{d}$. This was proven in Lemma \ref{lemma-compose-base-change-maps} (note that in the current case the morphisms $f', g'$ of that lemma are equal to $\text{id}_X$).