Any non-constant morphism $\Phi\colon X\to Y$ between projective smooth curves has a degree $d$ and the morphism $\Phi$ will be an isomorphism if and only if that degree is $1$.
The incredibly good news is that you can calculate that degree by just looking at the fibre $\Phi^{-1}(y)$ of $\Phi$ at just one point ( any point!) of $y\in Y$: the degree is the dimension $d=dim_k \Gamma (\Phi^{-1}(y),\mathcal O)$ .
It is not terribly difficult to explain what the right-hand side means, but this is not even necessary here: we have $d=1$ as soon as for some non-empty $V\subset Y$ the restricted morphism $\Phi^{-1}(V)\to V$ is an isomorphism.
Since this is true in your case, we are done.
Edit
A more elementary proof (maybe the one Hartshorne had in mind at the level of Chapter 1, before "degree" is introduced) would be to consider the inverse isomorphism $\psi:V\to U$, to complete it to a morphism $\bar \psi:\mathbb P^1\to \mathbb P^1 $ (just as you did for $\phi$) and realize that $\bar \psi $ is an inverse to $\bar \phi $, which is thus an isomorphism.
Regarding your first question, @Mohan posted a counterexample in the comments: $R = k[x,y,z]/(z^2 - x^2y)$ has an integral element $\frac{z}{x}$, which is not contained in $R$:
$$\left(\frac{z}{x}\right)^2 = \frac{z^2}{x^2} = \frac{x^2y}{x^2} = y,$$
so an integral relation would be $t^2 - y$.
To see that square-free is sufficient, and to answer your second question we need a Theorem about how integral extensions of integral domains and algebraic extensions of their quotient fields relate. This can be found in Matsumura's Commutative ring theory, p.65:
Theorem 9.2. Let $A$ be an integrally closed domain, $K$ the field of fractions of $A$, and $L$ an algebraic extension of $K$. Then an element $\alpha \in L$ is integral over $A$ if and only if its minimal polynomial over $K$ has all its coefficients in $A$.
Here $A = k[x_1,\dots,x_n]$ which is a UDF, hence integrally closed in $K = k(x_1,\dots,x_n)$, and $L = k(x_1,\dots,x_n)[z]/(z^2 - f)$.
Now we can see the formal reason why $f$ has to be square-free: We have to show that if $\alpha = g + hz$ is integral, both $g$ and $h$ belong to $A = k[x_1,\dots,x_n]$. But the Theorem only tells aus that the coefficients of the minimal polynomial $m_\alpha = X^2 - 2gX + (g^2 - h^2f)$ belong to $A$: $g \in A$, because $2 \neq 0$ in $k$ (good) and $g^2 - h^2f \in A$, so at least $h^2 f \in A$. Now if $f$ is square-free, it cannot compensate any denominators in $h$, because those appear twice. Hence $h \in A$ follows.
Best Answer
Question: "I know I can use 6.7 and 6.8 to extend the morphism f to such a ϕ. But I don't know how to prove it's surjective."
Answer: Let $C$ be an irreducible smooth projective curve over an algebraically closed field $k$ and let $K:=K(C)$ be the function field of $C$ with $x\in K$ a "non-constant" rational function. Let $P(t)\in k[t]$ and assume $P(x)=0$. Since $k$ is algebraically closed this implies $x\in k$ a contradiction, hence $x$ is "transcendental over $k$", meaning the subring $k[x] \subseteq K$ generated by $k$ and $x$ is isomorphic to a polynomial ring. You get an embedding $k \subseteq k(x) \subseteq K$ where $x$ behaves like an independent variable over $k$. This gives (by the proof of HH.I.6.12) a surjective morphism
$$\phi: C \rightarrow \mathbb{P}^1_k.$$
Since $C$ is projective and irreducible, the image is closed and irreducible and hence $Im(\phi)=\mathbb{P}^1_k$. The inverse image $\phi^{-1}(x)$ of a closed point $x\in \mathbb{P}^1_k$ is a strict closed subvariety of $C$ (since $\phi$ is non-constant) and since $C$ is irreducible it follows $\phi^{-1}(x)$ must be a finite set of points.