Consider category of $\mathbb{K}[x]$-modules. Let $\mathbb{K}$ be a trivial $\mathbb{K}[x]$-module, i.e. $x$ acts by zero. Easy to see that $\mathrm{Ext}^2 (\mathbb{K}, \mathbb{K}) = 0$. But there is exact sequence
$$0 \rightarrow \mathbb{K} \rightarrow \mathbb{K}[x]/(x^2) \rightarrow \mathbb{K}[x]/(x^2) \rightarrow \mathbb{K} \rightarrow 0.$$
It has to be equivalent to trivial one. It is due to this description of $\mathrm{Ext}^2$.
Question: How to construct this equivalence explicitly?
Best Answer
Late to the party, but anyway...
A length 2 exact sequence $$ 0 \to A \to B \to C \to D \to 0 $$ represents the zero class in $\mathrm{Ext}^2(D,A)$ if and only if we can fill in the hook diagram $$\require{AMScd}\begin{CD} @.@.@.0\\ @.@.@.@VVV\\ 0 @>>> A @>>> B @>>> I @>>> 0\\ @.@.@.@VVV\\ @.@.@.C\\ @.@.@.@VVV\\ @.@.@.D\\ @.@.@.@VVV\\ @.@.@.0 \end{CD}$$ where $I$ is the image of the map $C\to D$, to a pull-back/push-out diagram $$\require{AMScd}\begin{CD} @.@.0@.0\\ @.@.@VVV@VVV\\ 0 @>>> A @>>> B @>>> I @>>> 0\\ @.@|@VVV@VVV\\ 0 @>>> A @>>> E @>>> C @>>> 0\\ @.@.@VVV@VVV\\ @.@.D@=D\\ @.@.@VVV@VVV\\ @.@.0@.0 \end{CD}$$ for some $E$. In fact this works for any abelian (even exact) category, whether or not it has enough projectives or injectives.
In the example from the question this is clear: the image $I$ is just $\mathbb K$, and we can take for $E$ the module $\mathbb K[x]/(x^3)$, together with the obvious maps.