I can see that $S_3$ is not the direct sum of $A_3$ and $\{1,-1\}$, so that the sequence doesn't split. However, we were told that there still exist a map $\beta:\{1,-1\} \to S_3$ such that $g \beta=I_{\{1,-1\}}$, where $g:S_3 \to \{1,-1\}$ is the last map in the SES. And that there is no map $\alpha : S_3 \to A_3$ satisfying $\alpha f=I_{A_3}$.
Why if the sequence splits from the right does it not imply that the SES is split exact?
Also, how should one construct the map $\beta$ that then becomes a homomorphism and why doesn't no $\alpha$ exist?
If such an α existed, $S_3$ would be isomorphic to the direct product $A_3\times\{1,-1\}$ hence would be commutative since $A_3$ is a cyclic group of order $3$.
$\beta$ may be constructed with $$\beta(1)=\operatorname{id}=(),\quad\beta(-1)=(12)\quad\text{(or any transposition)}.$$