Let us consider a square in chain complexes over a field $k$ \begin{array}{ccc}A & \xrightarrow{f} & B \\ \downarrow{g} & & \downarrow{h} \\ C & \xrightarrow{k} & D\end{array}
If A is chain equivalent to a direct sum of chain complexes $\bigoplus_i A_i$ such that each of the summands $A_i$ is concentrated in exactly one degree but any two of $A_i$ s are in different degrees, then is it true that the square is equivalent (in the diagram category) to a direct sum of the squares with top left corner is $A_i$?
Any help will be appreciated.
I don’t think so. Colimits in functor categories are computed objectwise, so we would need not only $A$ but also $B,C,D$ to be direct sums with the same indices and the morphisms would be restricted.
What I think your question is motivated by is the following. Consider the square $$\tag{$\star$}\begin{array}{ccc} \bigoplus\limits_{i\in I} A_i & \rightarrow &B\\ \downarrow && \downarrow\\ C &\rightarrow &D \end{array}$$ By precomposition with the inclusion $A_j \hookrightarrow \bigoplus_i A_i$ we obtain a commutative square of the form $$\tag{$\star(j)$}\begin{array}{ccc} A_j& \rightarrow &B\\ \downarrow && \downarrow\\ C &\rightarrow &D \end{array}$$ for every $j$.
Conversely, if we have a square of the form $\star(j)$ for every $j$, denote the top and left morphisms by $b_j: A_j \rightarrow B$ and $c_j:A_j \rightarrow C$ respectively. By the universal property of the direct sum, they determine unique morphisms $b:\bigoplus_i A_i \rightarrow B$ and $c: \bigoplus_i A_i \rightarrow C$. Moreover uniqueness of the universal property implies that these morphisms yield a commutative square of the form $\star$.
So there is a bijective correspondence between squares of the form $\star$ and a collection of squares of the form $\star(j)$, but it is not just direct sums in the functor category.