Let $G$ be a finite group, and let $k$ be an algebraically closed field of characteristic $p$. An element $z\in H^n(G,k)$ can be represented by an $n$-fold extension of $k$ by $k$ in the category of $kG$-modules:
$$0\to k\to M_{n-1}\to\cdots\to M_0\to k\to 0$$
If $0\to k\to A\to k\to 0$ is a short exact sequence of $kG$-modules, then the image of the induced connecting homomorphism $H^n(G,k)\to H^{n+1}(G,k)$ on the extension above is given by just splicing:
$$0\to k\to A\to M_{n-1}\to\cdots\to M_0\to k\to 0$$
where the map $A\to M_{n-1}$ is the composite $A\to k\to M_{n-1}$.
The Bockstein is the map $\delta:H^n(G,\mathbb{F}_p)\to H^{n+1}(G,\mathbb{F}_p)$ induced by the short exact sequence
$$0\to \mathbb{Z}/p\mathbb{Z}\to\mathbb{Z}/p^2\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}\to 0$$
and since $H^n(G,k)\cong H^n(G,\mathbb{F}_p)\otimes_{\mathbb{F}_p}k$, $\delta$ extends $k$-linearly to a map $\delta': H^n(G,k)\to H^{n+1}(G,k)$. My question is:
What short exact sequence $0\to k\to A\to k\to0$ of $kG$-modules induces the map $\delta'$?
No such short exact sequence exists. Note that your "splicing" operation coincides with the cup product on $H^*(G,k)$, so if such a short exact sequence existed, then there would be a class $b\in H^1(G,k)$ such that $\delta'(x)=bx$ for all $x\in H^*(G,k)$. There usually does not exist any such class. For instance, let $G$ be cyclic of order $p$; then if $p$ is odd, $H^*(G,k)=k[x,y]/(y^2)$ with $|y|=1$ and $|x|=2$, and $\delta'$ is given by $\delta'(x^n)=0$ and $\delta'(x^ny)=x^{n+1}$. Clearly there is no element $b\in H^1(G,k)$ such that $\delta'$ is always multiplication by $b$. For $p=2$, we instead have $H^*(G,k)=k[y]$ with $|y|=1$ and $\delta'(y^{2n})=0$, $\delta'(y^{2n+1})=y^{2n+2}$. Again, $\delta'$ does not coincide with multiplication by any class in $H^1$.
(Note that when you replace $k$ by $\mathbb{F}_p$, $\delta$ is not induced by splicing with a short exact sequence of $kG$-modules, because $\mathbb{Z}/p^2\mathbb{Z}$ is not even a $k$-module! So there isn't really any particular reason to expect $\delta$ or $\delta'$ to come from splicing with a short exact sequence.)