Properties of a finite field extension of degree 2.

We have $x^q=x$ for all $x\in F$ and $x^{q^2}=x$ for all $x\in K$. Since $q$ is a power of $p$ and $p$ is characteristic of $K$, hence $\sigma\colon y\mapsto y^q$ is an automorphism of the field $K$. Note that $\sigma$ fixes all the points of $F$, and it can not fix any point of $K\setminus F$ (otherwise, the polynomial $X^q-X$ will have more than $q$ roots).

Now, $\sigma(\beta^{q+1})=\beta^{q(q+1)}=\beta^{q^2}\beta^{q}=\beta\beta^q=\beta^{q+1}$, thus $\beta^{q+1}$ is a fixed point of $\sigma$, it must be inside $F$, thats what you expected.

Here is the first part:

Let $\beta\in K$. We can assume that $\beta\ne0$ because $0^{1+q}=0\in F$.

If $\beta\ne0$, then, by Lagrange's theorem, we have $(\beta^{q+1})^{q-1}=\beta^{q^2-1}=1$.

Now, the roots of $x^{q-1}-1$ in $K$ are exactly the non-zero elements of $F$.

Therefore, $\beta^{q+1} \in F$.

For the second part:

The map $\beta \mapsto \beta^{q+1}$ is a group homomorphism $K^\times \to F^\times$. (The codomain is right by the first part.)

Its kernel are the roots of $x^{q+1}=1$. Since $K^\times$ is cyclic and $q+1$ divides $q^2-1$, the kernel is a subgroup of order $q+1$. Therefore, the image has order $(q^2-1)/(q+1)=q-1$, which is the order of $F^\times$. Thus, the map is surjective.