Ext groups due to Yoneda: why is this class zero? 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?
 A: 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.
A: 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 $\mathbb{K}$.
