Suppose there exists a map $\tilde h : K^0 \to K^{-1}$ such that $h = a^{-1} \circ \tilde h$ and such that $k^0 = \text{id} + d \circ \tilde h : K^0 \to K^0$ and $k^1 = \text{id} + \tilde h \circ d : K^{-1} \to K^{-1}$ are isomorphisms. Then we obtain a commutative diagram $$ \xymatrix{ 0 \ar[r] & \Ker(b^\bullet) \ar[r] \ar[d]_{c^\bullet} & K^\bullet \ar[r]_{b^\bullet} \ar[d]_{k^\bullet} & L^\bullet \ar[r] \ar[d]^{\text{id}} & 0 \\ 0 \ar[r] & \Ker(a^\bullet) \ar[r] & K^\bullet \ar[r]^{a^\bullet} & L^\bullet \ar[r] & 0 } $$ of complexes, where $c^\bullet$ is the induced isomorphism of kernels. Using Lemma \ref{lemma-det-ses-functorial} we see that $$ \xymatrix{ \det(\Ker(b^i)) \otimes \det(L^i) \ar[r] \ar[d]_{\det(c^i) \otimes 1} & \det(K^i) \ar[d]^{\det(k^i)} \\ \det(\Ker(a^i)) \otimes \det(L^i) \ar[r] & \det(K^i) } $$ commutes. Since $\det(c^\bullet)$ maps the canonical trivialization of $\det(\Ker(a^\bullet))$ to the same for $b^\bullet$ (Lemma \ref{lemma-canonical-element-well-defined}) we see that we conclude if (and only if) $$ \det(k^0) = \det(k^{-1}) $$ as elements of $R$ which follows from Lemma \ref{lemma-det-switch}. \medskip\noindent Suppose there exists a direct summand $U \subset K^{-1}$ such that both $a^{-1}|_U : U \to L^{-1}$ and $b^{-1}|_U : U \to L^{-1}$ are isomorphisms. Define $\tilde h$ as the composition of $h$ with the inverse of $a^{-1}|_U$. We claim that $\tilde h$ is a map as in the first paragraph of the proof. Namely, we have $h = a^{-1} \circ \tilde h$ by construction.  To show that $k^{-1} : K^{-1} \to K^{-1}$ is an isomorphism it suffices to show that it is surjective (Algebra, Lemma \ref{algebra-lemma-fun}). Let $u \in U$. We may choose $u' \in U$ such that $b^{-1}(u') = a^{-1}(u)$. Then $u = k^{-1}(u')$. Namely, both $u$ and $k^{-1}(u')$ are in $U$ and $a^{-1}(u) = a^{-1}(k^{-1}(u'))$ by a calculation\footnote{$a^{-1}(k^{-1}(u')) = a^{-1}(u') + a^{-1}(\tilde h(d(u'))) = a^{-1}(u') + h(d(u')) = b^{-1}(u') = a^{-1}(u)$} Since $a^{-1}|_U$ is an isomorphism we get the equality. Thus $U \subset \Im(k^{-1})$. On the other hand, if $x \in \Ker(a^{-1})$ then $x = k^{-1}(x) \bmod U$. Since $K^{-1} = \Ker(a^{-1}) + U$ we conclude $k^{-1}$ is surjective. Finally, we show that $k^0 : K^0 \to K^0$ is surjective. First, since $a^0 \circ k^0 = b^0$ we see that $a^0 \circ k^0$ is surjective. If $x \in \Ker(a^0)$, then $x = d(y)$ for some $y \in \Ker(a^{-1})$. We may write $y = k^{-1}(z)$ for some $z \in K^{-1}$ by the above. Then $x = k^0(d(z))$ and we conclude. \medskip\noindent Final step of the proof. It suffices to find $U$ as in the  preceding paragraph, but this may not always be possible. However, in order to show equality of two maps of $R$-modules, it suffices to do so after localization at primes of $R$. Hence we may assume $R$ is local. Then we get the following problem: suppose $$ \alpha, \beta : R^{\oplus n} \longrightarrow R^{\oplus m} $$ are two surjective $R$-linear maps. Find a direct summand $U \subset R^{\oplus n}$ such that both $\alpha|_U$ and $\beta|_U$ are isomorphisms. If $R$ is a field, this is possible by linear algebra. In general, one takes a solution over the residue field and lifts this to a solution over the local ring $R$. Some details omitted.