Field homomorphism into itself

Let $F=\mathbb Q(T)$. Then there is a homomorphism $F\to F$ given by $T\mapsto T^2$. It is not onto.

Edit: $\phi\colon F\to F$ is surjective because you can pick $\alpha\in F$ arbitrarily, then consider $S$ as in your quote, notice that $\phi$ is a permutation of the finite set $S$, hence there exists $\alpha'\in S$ with $\phi(\alpha')=\alpha$. In summary, for arbitrary $\alpha\in F$ there exists $\alpha'\in F$ such that $\phi(\alpha')=\alpha$.

Let $\theta\in F$. Restriction $\psi$ of $\varphi$ on $\mathbb{Q}(\theta)$ is a homomorphism of finite extensions of $\mathbb{Q}$. You've mentioned that it's injective. But $\psi$ is a linear mapping of finite dimensional $\mathbb{Q}$-spaces and so has a determinant. It does not vanish due to injectivity. Then $\psi$ has an inverse. So $(\theta\psi^{-1})\varphi = \theta$