Let $(M, K', \alpha)$ be an object of the target of the functor of the lemma. Here $\alpha : Lf^*M \to Li^*K'$ is an isomorphism which is adjoint to a map $\beta : M \to Rf_*Li^*K'$. Thus we obtain maps $$ Rj_*M \xrightarrow{Rj_*\beta} Rj_*Rf_*Li^*K' = Rf'_*Ri_*Li^*K' \leftarrow Rf'_*K' $$ where the arrow pointing left comes from $K' \to Ri_*Li^*K'$. Choose a distinguished triangle $$ M' \to Rj_*M \oplus Rf'_*K' \to Rj_*Rf_*Li^*K' \to M'[1] $$ in $D(\mathcal{O}_{Y'})$. The first arrow defines canonical maps $Lj^*M' \to M$ and $L(f')^*M' \to K'$ compatible with $\alpha$. Thus it suffices to show that the maps $Lj^*M' \to M$ and $L(f')^*M' \to K$ are isomorphisms. This we may check \'etale locally on $Y'$, hence we may assume $Y'$ is \'etale. \medskip\noindent Assume $Y'$ affine and $M \in D(\mathcal{O}_Y)$ and $K' \in D(\mathcal{O}_{X'})$ are pseudo-coherent. Say our pushout corresponds to the fibre product $$ \xymatrix{ B & B' \ar[l] \\ A \ar[u] & A' \ar[l] \ar[u] } $$ of rings where $B' \to B$ is surjective with locally nilpotent kernel $I$ (and hence $A' \to A$ is surjective with locally nilpotent kernel $I$ as well). The assumption on $M$ and $K'$ imply that $M$ comes from a pseudo-coherent object of $D(A)$ and $K'$ comes from a pseudo-coherent object of $D(B')$, see Derived Categories of Spaces, Lemmas \ref{spaces-perfect-lemma-pseudo-coherent}, \ref{spaces-perfect-lemma-derived-quasi-coherent-small-etale-site}, and \ref{spaces-perfect-lemma-descend-pseudo-coherent} and Derived Categories of Schemes, Lemma \ref{perfect-lemma-affine-compare-bounded} and \ref{perfect-lemma-pseudo-coherent-affine}. Moreover, pushforward and derived pullback agree with the corresponding operations on derived categories of modules, see Derived Categories of Spaces, Remark \ref{spaces-perfect-remark-match-total-direct-images} and Derived Categories of Schemes, Lemmas \ref{perfect-lemma-quasi-coherence-pushforward} and \ref{perfect-lemma-quasi-coherence-pullback}. This reduces us to the statement formulated in the next paragraph. (To be sure these references show the object $M'$ lies $D_\QCoh(\mathcal{O}_{Y'})$ as this is a triangulated subcategory of $D(\mathcal{O}_{Y'})$.) \medskip\noindent Given a diagram of rings as above and a triple $(M, K', \alpha)$ where $M \in D(A)$, $K' \in D(B')$ are pseudo-coherent and $\alpha : M \otimes_A^\mathbf{L} B \to K' \otimes_{B'}^\mathbf{L} B$ is an isomorphism suppose we have distinguished triangle $$ M' \to M \oplus K' \to K' \otimes_{B'}^\mathbf{L} B \to M'[1] $$ in $D(A')$. Goal: show that the induced maps $M' \otimes_{A'}^\mathbf{L} A \to M$ and $M' \otimes_{A'}^\mathbf{L} B' \to K'$ are isomorphisms. To do this, choose a bounded above complex $E^\bullet$ of finite free $A$-modules representing $M$. Since $(B', I)$ is a henselian pair (More on Algebra, Lemma \ref{more-algebra-lemma-locally-nilpotent-henselian}) with $B = B'/I$ we may apply More on Algebra, Lemma \ref{more-algebra-lemma-lift-complex-finite-projectives} to see that there exists a bounded above complex $P^\bullet$ of free $B'$-modules such that $\alpha$ is represented by an isomorphism $E^\bullet \otimes_A B \cong P^\bullet \otimes_{B'} B$. Then we can consider the short exact sequence $$ 0 \to L^\bullet \to E^\bullet \oplus P^\bullet \to P^\bullet \otimes_{B'} B \to 0 $$ of complexes of $B'$-modules. More on Algebra, Lemma \ref{more-algebra-lemma-finitely-presented-module-over-fibre-product} implies $L^\bullet$ is a bounded above complex of finite projective $A'$-modules (in fact it is rather easy to show directly that $L^n$ is finite free in our case) and that we have $L^\bullet \otimes_{A'} A = E^\bullet$ and $L^\bullet \otimes_{A'} B' = P^\bullet$. The short exact sequence gives a distinguished triangle $$ L^\bullet \to M \oplus K' \to K' \otimes_{B'}^\mathbf{L} B \to (L^\bullet)[1] $$ in $D(A')$ (Derived Categories, Section \ref{derived-section-canonical-delta-functor}) which is isomorphic to the given distinguished triangle by general properties of triangulated categories (Derived Categories, Section \ref{derived-section-elementary-results}). In other words, $L^\bullet$ represents $M'$ compatibly with the given maps. Thus the maps $M' \otimes_{A'}^\mathbf{L} A \to M$ and $M' \otimes_{A'}^\mathbf{L} B' \to K'$ are isomorphisms because we just saw that the corresponding thing is true for $L^\bullet$.