Let $\mathbb{Z}_n$ be the cyclic group of $n$ elements and $1$ the trivial group. I'm looking at the following extension:
$$1 \to \mathbb{Z}_2 \to \mathbb{Z}_4 \to \mathbb{Z}_2 \to 1$$
Because every action of $\mathbb{Z}_2$ on itself is trivial, the only split extension of $\mathbb{Z}_2$ by $\mathbb{Z}_2$ is the direct product $\mathbb{Z}_2 \times \mathbb{Z}_2 \cong V$ (the Klein group). But given that:
$$\mathbb{Z}_2 \cong \mathbb{Z}_4/\mathbb{Z}_2 = \{ (1 \bmod \mathbb{Z}_2), (x \bmod \mathbb{Z}_2)\} \text{ with } (x \bmod \mathbb{Z}_2)^2 = 1$$
A homomorphism $\psi: \mathbb{Z}_4/\mathbb{Z}_2 \to \mathbb{Z}_4$ that sends $(x \bmod \mathbb{Z}_2) \mapsto y$ (consider $y$ the generator of $\mathbb{Z}_4$) isn't a splitting of the above sequence? Because going through $\psi$ and then back through the projection sure is the identity mapping.
I'm sure I'm missing something deeper, because I also can't answer the general case: given that $G$ is a extension of $N$ (normal) by $K$, and that $K \cong G/N$, doesn't the injection $K \to G$ which maps $(x \bmod N) \mapsto x$ always splits the sequence? I'm sure not, but why?