Non-abelian extensions of the Klein four-group $K_4$ by $\mathbb{Z}_2$

Question

We know that we have eight extensions of $K_4$ by $\mathbb{Z}_2$ classified by $H^2_{grp}(K_4,\mathbb{Z}_2)$, see for example this link. Four of these extensions are abelian, and four of them are non-abelian. The $2$-cocyles for these extensions can be found for example in page 35 of this thesis. Considering the Cayley table for $K_4$, we know that $K_4$ is abelian. Now, if we want to define $2$-cocyles $f:K_4 \times K_4 \longrightarrow \mathbb{Z_2}$, we know that these cocycles need to satify $$f(a_1,a_2) + f(a_2,a_3) = f(a_1,a_3)$$ for all $a_1$, $a_2$ and $a_3$ in $K_4$. Here, we are using the definition of group cohomology on homogeneous cochains.
Every extension of $K_4$ by $\mathbb{Z}_2$ has underlying set $K_4 \times \mathbb{Z}_2$. The group operations on these extensions are given by $$(a_1,u) \cdot (a_2,v) = \bigl(a_1 a_2,uv f(a_1,a_2)\bigr).$$ Using the cocycle conditions, we have $f(a_1,a_1) = 0$ and $f(a_1,a_2) = - f(a_2,a_1)$ for all $a_1$ and $a_2$ in $K_4$. Considering $K_4$ as $\left\{e,a,b,c\right\}$ with group operation defined here, we have $$\begin{align*}(a_1,u) \cdot (a_2,v) & = (a_1a_2,uvf(a_1,a_2)) \\ & = (a_2a_1,vuf(a_2,a_1)) \\ & = (a_2,v) \cdot (a_1,u). \end{align*}$$ The second equality is because the values of $f$ are in $\mathbb{Z}_2$. So, my understanding is all these extensions obtained this way are abelian. Where are non-abelian extensions coming from using these $2$-cocycles? Where does my misunderstanding lie in?