Zero element of Ext$^n$ groups

If $A$ and $B$ are objects in the abelian category $\mathcal{A}$, the (equivalence class) of the exact sequence $$0\to A\to A\oplus B\to B\to 0$$ is the zero object in $\text{Ext}^{1}(B,A)$.

For $n>1$, the zero object of $\text{Ext}^{n}(B,A)$ is the (equivalence class) of the sequence

$$0 \to A\to A \to 0 \to\cdots\to 0\to B\to B\to 0$$ where the maps $A\to A$ and $B\to B$ are the identity.

The group structure on both is a bit fiddly. Consider the maps $\Delta:A\to A\oplus A$ given by $\begin{pmatrix}1 \\ 1\end{pmatrix}$ and $\nabla:A\oplus A\to A$ given by $\begin{pmatrix}1 & 1\end{pmatrix}$. The operation $+$ on $\text{Ext}^{i}(B,A)$ is $$\textbf{E}_{1}+\textbf{E}_{2} := \nabla (\textbf{E}_{1}\oplus\textbf{E}_{2})\Delta$$ where the direct sum is what you expect.

A proof that these form an abelian group can be found in Chapter 7 (called Extensions) of book Theory of Categories by Mitchell. It also shows that these coincide with the usual derived functor definition of Ext groups.

First one word of caution: for a general abelian category, even $\mathrm{Ext}^1(X,Y)$ can fail to be a set.

As people have pointed out above, the zero element in $\mathrm{Ext}^2(D,A)$ is represented by the four term exact sequence $$ 0 \to A \xrightarrow{1} A \xrightarrow{0} D \xrightarrow{1} D \to 0. $$ That still leaves the question of whether a given four term sequence is zero. For this there is the following nice answer. Given an exact sequence $$ \eta\colon 0 \to A \to B \to C \to D \to 0, $$ let $I$ be the image of the map $C\to D$. Then $\eta$ is zero in $\mathrm{Ext}^2(D,A)$ if and only if there exists an object $X$ fitting into an exact commutative diagram $$ \require{AMScd} \begin{CD} @.@. 0 @. 0\\ @.@. @VVV @VVV\\ 0 @>>> A @>>> B @>>> I @>>> 0\\ @. @| @VVV @VVV\\ 0 @>>> A @>>> X @>>> C @>>> 0\\ @.@. @VVV @VVV\\ @.@. D @= D\\ @.@. @VVV @VVV\\ @.@. 0 @. 0 \end{CD} $$ In fact, given another exact sequence $$ \eta' \colon 0 \to A \to B' \to C' \to D \to 0 $$ we have that $\eta=\eta'$ in $\mathrm{Ext}^2(D,A)$ if and only if there exists an object $X$ fitting into the exact commutative diagram $$ \require{AMScd} \begin{CD} @. 0 @. 0 @. 0\\ @. @VVV @VVV @VVV\\ 0 @>>> A @>>> B @>>> I @>>> 0\\ @. @VVV @VVV @VVV\\ 0 @>>> B' @>>> X @>>> C @>>> 0\\ @. @VVV @VVV @VVV\\ 0 @>>> I' @>>> C' @>>> D @>>> 0\\ @. @VVV @VVV @VVV\\ @. 0 @. 0 @. 0 \end{CD} $$