Why does this homological lemma hold? Let $A$ be a noetherian ring; let $C^{\boldsymbol\cdot}$ be a bounded above complex of flat $A$-modules in positive degrees, let $L^{\boldsymbol\cdot}$ be a bounded above complex of free $A$-modules in positive degrees, and let $g:L^{\boldsymbol\cdot}\rightarrow C^{\boldsymbol\cdot}$ be a morphism of complexes such that the induced map $h^i(L^{\boldsymbol\cdot})\rightarrow h^i(C^{\boldsymbol\cdot})$ is an isomorphism for each $i$. It is then claimed (Hartshorne III.12.3) that the map
$$h^i(L^{\boldsymbol\cdot}\otimes M)\rightarrow h^i(C^{\boldsymbol\cdot}\otimes M)$$
is an isomorphism for any $A$-module $M$.
Since $\otimes$ and $h^i$ commute with direct limits we reduce to $M$ finitely generated and write $$0\rightarrow R\rightarrow E\rightarrow M\rightarrow 0$$ for $E$ free f.g., $R$ the finitely-generated kernel. Since $L^{\boldsymbol\cdot}$ is a complex of free $A$-modules and $C^{\boldsymbol\cdot}$ is a complex of flat $A$-modules we get an exact, commutative diagram of complexes
$$\require{AMScd}
\begin{CD}
0@>>>L^{\boldsymbol\cdot}\otimes R@>>>L^{\boldsymbol\cdot}\otimes E@>>>L^{\boldsymbol\cdot}\otimes M@>>>0\\
@.@VVV@VVV@VVV\\
0@>>>C^{\boldsymbol\cdot}\otimes R@>>>C^{\boldsymbol\cdot}\otimes E@>>>C^{\boldsymbol\cdot}\otimes M@>>>0
\end{CD}
$$
Applying $h^i$, we get a commutative diagram of long-exact sequences
$$\require{AMScd}
\begin{CD}
\cdots@>>>h^i(L^{\boldsymbol\cdot}\otimes R)@>>>h^i(L^{\boldsymbol\cdot}\otimes E)@>>>h^i(L^{\boldsymbol\cdot}\otimes M)@>>>h^{i+1}(L^{\boldsymbol\cdot}\otimes R)@>>>h^{i+1}(L^{\boldsymbol\cdot}\otimes E)@>>>\cdots\\
@.@VVV@VVV@VVV@VVV@VVV\\
\cdots@>>>h^i(C^{\boldsymbol\cdot}\otimes R)@>>>h^i(C^{\boldsymbol\cdot}\otimes E)@>>>h^i(C^{\boldsymbol\cdot}\otimes M)@>>>h^{i+1}(C^{\boldsymbol\cdot}\otimes R)@>>>h^{i+1}(C^{\boldsymbol\cdot}\otimes E)@>>>\cdots
\end{CD}
$$
Since the result holds for $i+1$ by induction, and for $E$, since $E$ is free and $h^i(L^{\boldsymbol\cdot})\rightarrow h^i(C^{\boldsymbol\cdot})$ is an isomorphism, one deduces that the five vertical arrows in the diagram above are (?,iso,?,iso,iso). It follows that the middle map is an epimorphism from one of the four-lemmas but it is concluded ('from the subtle 5-lemma') that the middle map is in fact an isomorphism. To apply the 5-lemma I know, we also need that the leftmost vertical arrow is an epimorphism. I don't see why this is.
 A: Let me first give some names to your morphisms:
$$\require{AMScd}
\begin{CD}
\cdots@>>>h^i(L^{\bullet}\otimes R)@>>>h^i(L^{\bullet}\otimes E)@>>>h^i(L^{\bullet}\otimes M)@>>>h^{i+1}(L^{\bullet}\otimes R)@>>>h^{i+1}(L^{\bullet}\otimes E)@>>>\cdots\\
@.@VV{\varphi^i_R}V@V\cong V{\varphi^i_E}V@VV{\varphi^i_M}V@V\cong V{\varphi^{i+1}_R}V@V\cong V{\varphi^{i+1}_E}V\\
\cdots@>>>h^i(C^{\bullet}\otimes R)@>>>h^i(C^{\bullet}\otimes E)@>>>h^i(C^{\bullet}\otimes M)@>>>h^{i+1}(C^{\bullet}\otimes R)@>>>h^{i+1}(C^{\bullet}\otimes E)@>>>\cdots
\end{CD}
$$
As you say, $\varphi^{i+1}_R$ is an isomorphism by induction, and $\varphi^{i}_E,\varphi^{i+1}_E$ are isomorphisms since $E$ is free.
Now we can say the following:


*

*By the 4-lemma, $\varphi^i_M$ is an epimorphism; that is, we have shown that
$$\varphi^i_M\colon h^i(L^\bullet \otimes M) \to h^i(C^\bullet \otimes M)$$
is an epimorphism for any finitely-generated $M$.

*Now replacing $M$ with $R$ in Step 1, we also have that $\varphi^i_R$ is an epimorphism.

*By the 5-lemma, $\varphi^i_M$ is then an isomorphism.

