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?
Take the projective resolution $$0\to0\to\mathbb{K}[x]\to\mathbb{K}[x]\to\mathbb{K}\to0$$ of $\mathbb{K}$, which you can regard as a length $2$ extension of $\mathbb{K}$ by $0$, and take the direct sum with $$0\to\mathbb{K}\to\mathbb{K}\to0\to0\to0$$ to get $$0\to\mathbb{K}\to\mathbb{K}\oplus\mathbb{K}[x]\to\mathbb{K}[x]\to\mathbb{K}\to0,$$ which is a length $2$ extension of $\mathbb{K}$ by $\mathbb{K}$.
But this extension has a map to both your extension $$0\to\mathbb{K}\to\mathbb{K}[x]/(x^2)\to\mathbb{K}[x]/(x^2)\to\mathbb{K}\to0$$ and to the trivial extension $$0\to\mathbb{K}\to\mathbb{K}\stackrel{0}{\to}\mathbb{K}\to\mathbb{K}\to0,$$ so all three extensions are equivalent. In fact, it has a map to every length $2$ extension of $\mathbb{K}$ by $\mathbb{K}$.