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.
$\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\Ext}{Ext}$I must have a different edition than you do, here's the statement of Theorem 3.4.3 that I read:
Theorem 3.4.3. Given two $R$-modules $A$ and $B$, the mapping $\Theta : \xi \mapsto \partial(\operatorname{id}_A)$ establishes a 1-1 correspondence
$$\{\text{equivalence classes of extensions of } A \text{ by } B \} \leftrightarrow \Ext^1(A,B)$$
in which the split extension corresponds to the element $0 \in \Ext^1(A,B)$.
Proof Fix an exact sequence $0 \to M \xrightarrow{j} P \to A \to 0$ with $P$ projective. Applying $\Hom(-, B)$ yields an exact sequence
$$\Hom(P, B) \to \Hom(M,B) \xrightarrow{\partial} \Ext^1(A,B) \to 0.$$
[...]
Now the mapping $\Theta$ is actually explained above the statement of the theorem. $\xi$ is an exact sequence, of the form $0 \to B \to X \to A \to 0$. If you apply the theory of derived functors to compute $\mathbb{R}\Hom(A,-)$, you get a long exact sequence:
$$0 \to \Hom(A,B) \to \Hom(A,X) \to \Hom(A,A) \xrightarrow{\partial} \Ext^1(A,B) \to \Ext^1(A,X) \to \dots$$
where $\partial$ is the connecting morphism, for this particular exact sequence $\xi$. And the image of $\operatorname{id}_A$ under this $\partial$ is $\Theta(\xi)$.
But in the proof, Weibel takes also another exact sequence, $0 \to M \xrightarrow{j} P \to A \to 0$, and applies again derived functors, but on the other side of the $\Hom$! (So you're computing $\mathbb{R}\Hom(-, B)$.) You get a long exact sequence:
$$0 \to \Hom(A,B) \to \Hom(P,B) \to \Hom(M,B) \xrightarrow{\partial} \Ext^1(A,B) \to \Ext^1(P,B) \to \dots$$
there is also a connecting morphism. We can call it $\partial$ again, but it's not the same $\partial$ as before, it's just that connecting homomorphisms are called $\partial$ in general (or $\delta$, or some other variant on the letter "d"). In general the context is enough to determine what $\partial$ we are talking about.
It's like how the differential of a chain complex is usually simply called "$d$", even if there are multiple chain complexes that we are considering... You could call the first connecting morphism $\partial_\xi^A$ or something like that, and find an appropriate notation for the second one, but carrying all these subscripts and superscripts around becomes annoying very quickly. (And don't forget that this is only the first connecting morphism, there are higher Ext groups! So it should be something like $\partial_{\xi, A}^1$...)
Bonus: In the proof, $P$ is projective, which is why $\Ext^1(P,B) = 0$ and why Weibel omitted it in the long exact sequence.
Best Answer
You can just splice the short exact sequence at the end of the long extension. That is, given a long extension $$0 \to E \to Z_{i-1} \to \cdots \to Z_0 \to A \to 0,$$ the connecting homomorphism associated to the short exact sequence $0 \to A \to B \to C \to 0$ takes this to the long(er) extension $$0 \to E \to Z_{i-1} \to \cdots \to Z_0 \xrightarrow{(*)} B \to C \to 0,$$ where the only new map is $(*)$, which is defined as the composite $Z_0 \to A \to B$.
From this description, we see that the extensions in the kernel of $\DeclareMathOperator{\Ext}{Ext} \Ext^2(C, E) \to \Ext^2(B, E)$ are those of the form $$0 \to E \to Z \to B \to C \to 0.$$ To see that these extensions are indeed in the kernel, pull it back along $B \to C$ to get $$0 \to E \to {?} \to B \times_C B \to B \to 0.$$ Since $B \times_C B \cong B \oplus \ker(B \to C) \cong B \oplus \operatorname{im}(Z \to B)$, we see that this extension must be $$0 \to E \to Z \to B \oplus \operatorname{im}(Z \to B) \to B \to 0,$$ which is the direct sum of $$0 \to E \to Z \to \operatorname{im}(Z \to B) \to 0 \to 0$$ with $$0 \to 0 \to 0 \to B \xrightarrow{1} B \to 0.$$ The former is equivalent to $0 \to E \xrightarrow{1} E \to 0 \to 0$, so the pullback is equivalent to $$0 \to E \xrightarrow{1} E \xrightarrow{0} B \xrightarrow{1} B \to 0,$$ which is the zero in the $\Ext$ group.