Obstruction to splitting an object in derived category into a sum of two-term complexes Let $\mathcal{A}$ be an abelian category, and $D$ its bounded derived category. An object $M \in D$ may be described as a list of cohomology objects $H^i = H^i(M)$ together with some complicated glueing data.
I am interested only in the case when $\mathcal{A}$ has homological dimension two. For example, $\mathcal{A}$ can be a category of coherent sheaves on a smooth surface. In this case the glueing data amounts to a collection of classes $\xi_i \in \mathrm{Ext}^2(H^i, H^{i-1})$ between each pair of adjacent cohomology objects, with no restrictions on the choices.
By definition, an object $M \in D$ is quasiisomorphic to a direct sum of complexes concentrated in a single degree (i.e., shifts of objects from $\mathcal{A}$) if and only if each $\xi_i$ vanishes. 
Similarly, some objects in $D$ are quasiisomorphic to direct sums of complexes concentrated in two adjacent degrees. Is it possible to characterize this property by vanishing of some obstructions built in terms of the presentation of an object as a collection $\{ (H_i, \xi_i) \}_{i \in \mathbb{Z}}$ above?
 A: As pointed out in nikola karabatic's answer, a decomposition of M as a direct sum of two-term complexes induces decompositions $H^i=H^i_a\oplus H^i_b$ for each $i$. These have the property that $\xi_i\in\text{Ext}^2(H^i_a,H^{i-1}_b)$, where $\text{Ext}^2(H^i_a,H^{i-1}_b)$ is regarded as a direct summand of $\text{Ext}^2(H^i,H^{i-1})$ in the obvious way. Or equivalently, there are vanishing obstructions in the other three summands of $\text{Ext}^2(H^i,H^{i-1})$.
Conversely, if there are such decompositions of the $H^i$, then the reconstruction of $M$ from the data $\{(H^i,\xi_i)\}_{i\in\mathbb{Z}}$ gives a direct sum of two term complexes with cohomology $H^i_a$ in degree $i$ and $H^{i-1}_b$ in degree $i-1$.
A: This is not a complete answer, but too long for a comment.
A splitting into a sum of two chain complexes yields a splitting of every $H_i$ into two parts $H_{i}^1\oplus H_{i}^2$, coming from the complex in degrees $i-1,i$ and $i, i+1$ respectively. This splitting should be part of the data defining the obstruction (otherwise I see no chance to do so).
Given this data, consider some fixed index $i$. We consider the distinguished triangle
$$\tau_{[i-1,i]}M \rightarrow \tau_{[i-1,i+1]}M\rightarrow H_{i+1}M[-i-1]\rightarrow \tau_{[i-1,i]}M[1].$$
Here $\tau_{[i-1,i]}M=\tau_{\ge i}\tau_{\le i} M$ and $\tau_{\le n}$ denotes the canonical truncation.
The cohomology splits as $H_{i+1}M=H_{i+1}^1\oplus H_{i+1}^2$. Let $M'$ be the preimage of $H_{i+1}^1$ in $\tau_{[i-1,i+1]}M$, then we have a triangle
$$\tau_{[i-1,i]}M \rightarrow M' \rightarrow H_{i+1}^1  M[-i-1]\rightarrow \tau_{[i-1,i]}M[1]$$
and want to know whether this splits. For this splitting, there is an obstruction $\eta_i$ in $\mathrm{Hom}_{D(\mathcal{A})}(H_{i+1}^1  M[-i-1], \tau_{[i-1,i]}M[1])$ given as the image of the identity in the long exact sequence
$$ \ldots\rightarrow \mathrm{Hom}_{D(\mathcal{A})}(H_{i+1}^1  M[-i-1],M')\rightarrow \mathrm{Hom}_{D(\mathcal{A})}(H_{i+1}^1  M[-i-1], H_{i+1}^1  M[-i-1]) \rightarrow  \mathrm{Hom}_{D(\mathcal{A})}(H_{i+1}^1  M[-i-1], \tau_{[i-1,i]}M[1])\rightarrow \ldots $$
The complex $\tau_{[i-1,i]}M[1]$ is given by your $\xi_i$.
Similarly, you get an obstruction $\mu_i$ by considering $\tau_{[i-2,i]}M$. 
Upshot: The obstructions $\eta_i$ and $\mu_i$ constructed above, depending on a splitting of the cohomology, certainly vanish if $M$ splits as a sum of two-term complexes. I don't know whether the converse is true.
