By the construction in Remark \ref{remark-cup-product} we see that going around the diagram clockwise the map $$ f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} f_*\mathcal{M}^\bullet  \longrightarrow Rf_*\text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) $$ is adjoint to the map \begin{align*} Lf^*(f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} f_*\mathcal{M}^\bullet) & = Lf^*f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} Lf^*f_*\mathcal{M}^\bullet \\ & \to Lf^*Rf_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} Lf^*Rf_*\mathcal{M}^\bullet \\ & \to \mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} \mathcal{M}^\bullet \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) \end{align*} By Lemma \ref{lemma-adjoints-push-pull-compatibility} this is also equal to \begin{align*} Lf^*(f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} f_*\mathcal{M}^\bullet) & = Lf^*f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} Lf^*f_*\mathcal{M}^\bullet \\ & \to f^*f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} f^*f_*\mathcal{M}^\bullet \\ & \to \mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} \mathcal{M}^\bullet \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) \end{align*} Going around anti-clockwise we obtain the map adjoint to the map \begin{align*} Lf^*(f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} f_*\mathcal{M}^\bullet) & \to Lf^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y} f_*\mathcal{M}^\bullet) \\ & \to Lf^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) \\ & \to Lf^*Rf_*\text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) \end{align*} By Lemma \ref{lemma-adjoints-push-pull-compatibility} this is also equal to \begin{align*} Lf^*(f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} f_*\mathcal{M}^\bullet) & \to Lf^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y} f_*\mathcal{M}^\bullet) \\ & \to Lf^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) \\ & \to f^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) \end{align*} Now the proof is finished by a contemplation of the diagram $$ \xymatrix{ Lf^*(f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y}^\mathbf{L} f_*\mathcal{M}^\bullet) \ar[d] \ar[rr] & & Lf^*f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_X}^\mathbf{L} Lf^*f_*\mathcal{M}^\bullet \ar[d] \\ Lf^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y} f_*\mathcal{M}^\bullet) \ar[d]_{naive} \ar[r] & f^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_Y} f_*\mathcal{M}^\bullet) \ar[ldd]^{naive} \ar[dd] & f^*f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_X}^\mathbf{L} f^*f_*\mathcal{M}^\bullet \ar[dd] \ar[ldd] \\ Lf^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) \ar[d] \\ f^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) \ar[rd] & \text{Tot}(f^*f_*\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} f^*f_*\mathcal{M}^\bullet) \ar[d] & \mathcal{K}^\bullet \otimes_{\mathcal{O}_X}^\mathbf{L} \mathcal{M}^\bullet \ar[ld] \\ & \text{Tot}(\mathcal{K}^\bullet \otimes_{\mathcal{O}_X} \mathcal{M}^\bullet) } $$ All of the polygons in this diagram commute. The top one commutes by Lemma \ref{lemma-tensor-pull-compatibility}. The square with the two naive cup products commutes because $Lf^* \to f^*$ is functorial in the complex of modules. Similarly with the square involving the two maps $\mathcal{A}^\bullet \otimes^\mathbf{L} \mathcal{B}^\bullet \to \text{Tot}(\mathcal{A}^\bullet \otimes \mathcal{B}^\bullet)$. Finally, the commutativity of the remaining square is true on the level of complexes and may be viewed as the definiton of the naive cup product (by the adjointness of $f^*$ and $f_*$). The proof is finished because going around the diagram on the outside are the two maps given above.