I didn't feel MO was the best place to ask this question, so apologies for this, but when I asked it at https://math.stackexchange.com/questions/2297837/why-is-this-cubic-polynomial-generic-for-cyclic-field-extensions, I didn't get enough information. I would really like to understand this example, so I will try to streamline the question and ask it here.

I am trying to understand the circumstances in which a finite cover of $\mathbb{P}^1_K$ of Galois group $G$ can be twisted to contain a field extension $L/K$ of group $G$. This construction comes from p1 of Serre's Topics in Galois Theory.

Suppose we have a field $K$, which I would like to think of as $\mathbb{Q}$, and take the curve $Y = \mathbb{P}^1_K$ and a finite subgroup $G \subset \mathrm{Aut}(Y)$, where $G\cong \mathbb{Z}/3\mathbb{Z}$. Now treat $Y$ as a finite branched cover of $\mathbb{P}^1$ via the quotient $Y \to Y/G$.

If $L/K$ is a Galois extension also with group $G$, then we get a map $\phi:G_K \to G \to \mathrm{Aut}(Y)$, which we can view as a 1-cocycle $\Big($because with trivial action $H^1(G_K,\mathrm{Aut}(Y)) = \mathrm{Hom}(G_K,\mathrm{Aut}(Y))$, right?$\Big)$.

(1) Why is the extension $L/K$ given by a rational point on $\mathbb{P}^1_K/G$ if and only if the twist of $Y$ by this cocycle has a rational point not invariant by G?

(2) How explicitly can I understand the twist of $Y$ by a cocycle? Can I get equations for it?

Thanks.