I already understand t.gunn's proof (below) but got me thinking how did he come up with the idea? I mean, why did he know that the roots of the minimal polynomial had to be elements from here $G\cdot\alpha$ (the orbit of $\alpha)?$
Proposition: If $\mathbf K:\mathbf F<\infty,$ and is Galois, then it is normal and separable.
Proof: (summary of t.gunn's proof :)
Let $\alpha\in\mathbf K.$
The minimal polynomial of $\alpha$ is
$$ f_\alpha(x) := \prod_{\beta \in G \cdot \alpha} (x - \beta). $$ Indeed:
First, note that $f_\alpha(\alpha) = 0$, which follows since $\alpha = \operatorname{id}(\alpha) \in G \cdot \alpha$.
Second, note that $f_\alpha \in \mathbf{F}[x]$,
Third note that $f_\alpha$ is minimal. Indeed if $f(\alpha) = 0$ then $f(\sigma(\alpha)) = \sigma(f(\alpha)) = \sigma(0) = 0$ for all $\sigma \in G$. Thus $\sigma(\alpha)$ is a root for all $\sigma \in G$. Thus $f_\alpha \mid f$.
Finally, we note that $f_\alpha$ splits over $\mathbf{K}$ and is separable, by construction.
Let $K/F$ be a finite Galois extension, and let $G = \textrm{Gal}(K/F)$. Let $\alpha \in K$, and let $f$ be the minimal polynomial of $\alpha$ over $F$. Let $\Phi$ be the set of roots of $f$. Since $K/F$ is normal, $\Phi$ is contained in $K$. Since $K/F$ is separable, the roots of $f$ are distinct. Thus
$$f(x) = \prod\limits_{\beta \in \Phi} (x - \beta)$$
If I understand your question, what you want to know is why the set $\Phi$ is equal to the set $G.\alpha = \{ \sigma(\alpha) : \sigma \in G \}$.
First, $G.\alpha$ is contained in $\Phi$. This is easy to see, because if $\sigma \in G$, then $\sigma$ fixes the elements of $F$, hence the coefficients of $f$.
The converse inclusion is more difficult. You can prove it using the two equivalent definitions of what it means for a field extension to be normal:
Proposition: Let $E$ be an algebraic extension of a field $F$. The following are equivalent:
(i): If $a \in E$, then all the roots of the minimal polynomial of $a$ over $F$ are also in $E$.
(ii): If $\Omega$ is a field containing $E$, and $\phi: E \rightarrow \Omega$ is a homomorphism which fixes $F$ (pointwise), then $\phi$ is actually just an automorphism of $E$.
Now let $\beta$ be another root of $f$. We want to show that $\beta = \sigma(\alpha)$ for some $\sigma \in G$. There is a unique isomorphism of fields $\phi: F(\alpha) \rightarrow F(\beta)$ which fixes $F$ pointwise and sends $\alpha$ to $\beta$.
Fix an algebraic closure $\overline{F}$ of $F$ containing $E$. The isomorphism extension theorem (https://en.wikipedia.org/wiki/Isomorphism_extension_theorem) tells you that $\phi$ extends (in general, nonuniquely!) to homomorphism $\tilde{\phi}$ of $K$ into $\overline{F}$. This can be proved using Zorn's lemma.
But since $K/F$ is a normal extension, $\tilde{\phi}$ is just an automorphism of $K$ fixing $F$. In other words, $\tilde{\phi} \in G$. And $\tilde{\phi}(\alpha) = \phi(\alpha) = \beta$, as required.