The ``only if'' direction is implied by assumption (2). On the other hand, if $K'$ has a unique nonzero cohomology sheaf, then the ``if'' direction follows from assumption (1). In general we will use an induction argument to prove the ``if'' direction. Let us say an object $K'$ of $D^+(\mathcal{O}_{\tau'})$ satisfies (P) if the maps $c^{K'}_{X_\alpha, Z_\alpha, Y_\alpha, E_\alpha}$ are isomorphisms for all $\alpha \in A$. \medskip\noindent Namely, let $K'$ be an object of $D^+(\mathcal{O}_{\tau'})$ satisfying (P). Choose a bounded below complex ${\mathcal{K}'}^\bullet$ of sheaves of $\mathcal{O}_{\tau'}$-modules representing $K'$. We will show by induction on $n$ that we may assume for $p \leq n$ we have $(\mathcal{K}')^p = \epsilon_*\mathcal{J}^p$ for some injective sheaf $\mathcal{J}^p$ of $\mathcal{O}_{\tau}$-modules. The assertion is true for $n \ll 0$ because $(\mathcal{K}')^\bullet$ is bounded below. \medskip\noindent Induction step. Assume we have $(\mathcal{K}')^p = \epsilon_*\mathcal{J}^p$ for some injective sheaves $\mathcal{J}^p$ of $\mathcal{O}_\tau$-modules for $p \leq n$. Denote $\mathcal{J}^\bullet$ the bounded complex of injective $\mathcal{O}_\tau$-modules made from these sheaves and the maps between them. Consider the short exact sequence of complexes $$ 0 \to \sigma_{\geq n + 1}(\mathcal{K}')^\bullet \to (\mathcal{K}')^\bullet \to \epsilon_*\mathcal{J}^\bullet \to 0 $$ where $\sigma_{\geq n + 1}$ denotes the ``stupid'' truncation. By assumption (2) the object $\epsilon_*\mathcal{J}^\bullet$ of $D(\mathcal{O}_{\tau'})$ satisfies (P). By Lemma \ref{lemma-two-out-of-three-blow-up-square} we conclude that $\sigma_{\geq n + 1}(\mathcal{K}')^\bullet$ satisfies (P). We conclude that for $\alpha \in A$ the sequence $$ \begin{matrix} 0 \\ \downarrow \\ H^{n + 1}_{\tau'}(X_\alpha, \sigma_{\geq n + 1}(\mathcal{K}')^\bullet) \\ \downarrow \\ H^{n + 1}_{\tau'}(Z_\alpha, \sigma_{\geq n + 1}(\mathcal{K}')^\bullet) \oplus H^{n + 1}_{\tau'}(Y_\alpha, \sigma_{\geq n + 1}(\mathcal{K}')^\bullet) \\ \downarrow \\ H^{n + 1}_{\tau'}(E_\alpha, \sigma_{\geq n + 1}(\mathcal{K}')^\bullet) \end{matrix} $$ is exact by the distinguished triangle of Lemma \ref{lemma-c-square} and the fact that $\sigma_{\geq n + 1}(\mathcal{K}')^\bullet$ has vanishing cohomology over $E_\alpha$ in degrees $< n + 1$. We conclude that $$ \mathcal{F}' = \Ker((\mathcal{K}')^{n + 1} \to (\mathcal{K}')^{n + 2}) $$ is a $\tau$-sheaf by assumption (1) because the cohomology groups above evaluate to $\mathcal{F}'(X_\alpha)$, $\mathcal{F}'(Z_\alpha) \oplus \mathcal{F}'(Y_\alpha)$, and $\mathcal{F}'(E_\alpha)$. Thus we may choose an injective $\mathcal{O}_\tau$-module $\mathcal{J}^{n + 1}$ and an injection $\mathcal{F}' \to \epsilon_*\mathcal{J}^{n + 1}$. Since $\epsilon_*\mathcal{J}^{n + 1}$ is also an injective $\mathcal{O}_{\tau'}$-module (Lemma \ref{lemma-pushforward-injective-flat}) we can extend $\mathcal{F}' \to \epsilon_*\mathcal{J}^{n + 1}$ to a map $(\mathcal{K}')^{n + 1} \to \epsilon_*\mathcal{J}^{n + 1}$. Then the complex $(\mathcal{K}')^\bullet$ is quasi-isomorphic to the complex $$ \ldots \to \epsilon_*\mathcal{J}^n \to \epsilon_*\mathcal{J}^{n + 1} \to \frac{\epsilon_*\mathcal{J}^{n + 1} \oplus (\mathcal{K}')^{n + 2}}{(\mathcal{K}')^{n + 1}} \to (\mathcal{K}')^{n + 3} \to \ldots $$ This finishes the induction step. \medskip\noindent The induction procedure described above actually produces a sequence of quasi-isomorphisms of complexes $$ (\mathcal{K}')^\bullet \to (\mathcal{K}'_{n_0})^\bullet \to (\mathcal{K}'_{n_0 + 1})^\bullet \to (\mathcal{K}'_{n_0 + 2})^\bullet \to \ldots $$ where $(\mathcal{K}'_n)^\bullet \to (\mathcal{K}'_{n + 1})^\bullet$ is an isomorphism in degrees $\leq n$ and such that $(\mathcal{K}'_n)^p = \epsilon_*\mathcal{J}^p$ for $p \leq n$. Taking the ``limit'' of these maps therefore gives a quasi-isomorphism $(\mathcal{K}')^\bullet \to \epsilon_*\mathcal{J}^\bullet$ which proves the lemma.