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 $$\tau_{[i-1,i]}M \rightarrow \tau_{[i-1,i+1]}M\rightarrow H_{i+1}M\rightarrow \tau_{[i-1,i]}M[1].$$ 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 $$\tau_{[i-1,i]}M \rightarrow M' \rightarrow H_{i+1}^1 M\rightarrow \tau_{[i-1,i]}M[1]$$ and want to know whether this splits. For this splitting, there is an obstruction $\eta_i$ in $\mathrm{Hom}_{D(\mathcal{A})}(H_{i+1}^1 M, \tau_{[i-1,i]}M[1])$ given as the image of the identity in the long exact sequence $$ \ldots\rightarrow \mathrm{Hom}_{D(\mathcal{A})}(H_{i+1}^1 M,M')\rightarrow \mathrm{Hom}_{D(\mathcal{A})}(H_{i+1}^1 M, H_{i+1}^1 M) \rightarrow \mathrm{Hom}_{D(\mathcal{A})}(H_{i+1}^1 M, \tau_{[i-1,i]}M[1])\rightarrow \ldots $$ 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.

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.