Observe that (1) is \'etale local on $X'$. Working through formal implications of what is known, we see that it suffices to prove condition (1) of this lemma is equivalent to condition (1) of Derived Categories of Schemes, Lemma \ref{perfect-lemma-single-complex-base-change-condition} if $X, Y, Y', X'$ are representable by schemes $X_0, Y_0, Y'_0, X'_0$. Denote $f_0, g_0, g'_0, f'_0$ the morphisms between these schemes corresponding to $f, g, g', f'$. We may assume $K = \epsilon^*K_0$ and $K' = \epsilon^*K'_0$ for some objects $K_0 \in D_\QCoh(\mathcal{O}_{X_0})$ and $K'_0 \in D_\QCoh(\mathcal{O}_{X'_0})$, see Lemma \ref{lemma-derived-quasi-coherent-small-etale-site}. Moreover, the map $Lg^*K \to K'$ is the pullback of a map $L(g_0)^*K_0 \to K'_0$ with notation as in Remark \ref{remark-match-total-direct-images}. Recall that $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{X, x}$ (Properties of Spaces, Lemma \ref{spaces-properties-lemma-describe-etale-local-ring}) and that we have $$ K_{\overline{x}} = K_{0, x} \otimes_{\mathcal{O}_{X, x}}^\mathbf{L} \mathcal{O}_{X, \overline{x}} \quad\text{and}\quad K'_{\overline{x}'} = K'_{0, x'} \otimes_{\mathcal{O}_{X', x'}}^\mathbf{L} \mathcal{O}_{X', \overline{x}'} $$ (akin to Properties of Spaces, Lemma \ref{spaces-properties-lemma-stalk-quasi-coherent}). Consider the commutative diagram $$ \xymatrix{ H^i(K_{\overline{x}} \otimes_{\mathcal{O}_{Y, \overline{y}}}^\mathbf{L} \mathcal{O}_{Y', \overline{y}'}) \otimes_{(\mathcal{O}_{X, \overline{x}} \otimes_{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{Y', \overline{y}'})} \mathcal{O}_{X', \overline{x}'} \ar[r] & H^i(K'_{\overline{x}'}) \\ H^i(K_{0, x} \otimes_{\mathcal{O}_{Y, y}}^\mathbf{L} \mathcal{O}_{Y', y'}) \otimes_{(\mathcal{O}_{X, x} \otimes_{\mathcal{O}_{Y, y}} \mathcal{O}_{Y', y'})} \mathcal{O}_{X', x'} \ar[u] \ar[r] & H^i(K'_{0, x'}) \ar[u] } $$ We have to show that the lower horizontal arrow is an isomorphism if and only if the upper horizontal arrow is an isomorphism. Since $\mathcal{O}_{X', x'} \to \mathcal{O}_{X', \overline{x}'}$ is faithfully flat (More on Algebra, Lemma \ref{more-algebra-lemma-dumb-properties-henselization}) it suffices to show that the top arrow is the base change of the bottom arrow by this map. This follows immediately from the relationships between stalks given above for the objects on the right. For the objects on the left it suffices to show that \begin{align*} & H^i\left( (K_{0, x} \otimes_{\mathcal{O}_{X, x}}^\mathbf{L} \mathcal{O}_{X, \overline{x}}) \otimes_{\mathcal{O}_{Y, \overline{y}}}^\mathbf{L} \mathcal{O}_{Y', \overline{y}'}\right) \\ & = H^i(K_{0, x} \otimes_{\mathcal{O}_{Y, y}}^\mathbf{L} \mathcal{O}_{Y', y'}) \otimes_{(\mathcal{O}_{X, x} \otimes_{\mathcal{O}_{Y, y}} \mathcal{O}_{Y', y'})} (\mathcal{O}_{X, \overline{x}} \otimes_{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{Y', \overline{y}'}) \end{align*} This follows from More on Algebra, Lemma \ref{more-algebra-lemma-lemma-tor-independent-flat-compare}. The flatness assumptions of this lemma hold by what was said above as well as Algebra, Lemma \ref{algebra-lemma-strictly-henselian-functorial-improve} implying that $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{X, x} \otimes_{\mathcal{O}_{Y, y}} \mathcal{O}_{Y, \overline{y}}$ and that $\mathcal{O}_{Y', \overline{y}'}$ is the strict henselization of $\mathcal{O}_{Y', y'} \otimes_{\mathcal{O}_{Y, y}} \mathcal{O}_{Y, \overline{y}}$.