This is a follow-up question to this one: Rational functions on curve
In that setting, assume $X=\mathbb{P}^1$ and $f$ an isomorphism, so that we are looking at automorphisms of $\mathbb{P}^1$. Denoting the coordinate ring of $\mathbb{A}^1$ as $k[x]$, we have that the rational functions on $\mathbb{P}^1$, denoted by $K(\mathbb{P}^1)$, is the field of fractions of $k[x] = \mathcal{O}_{\mathbb{P}^1}(\mathbb{A}^1)$. I am trying to show that $f^*:K(\mathbb{P}^1) \rightarrow K(\mathbb{P}^1)$ induced by $f$, sends $x$ to $f^{\vee}$ (defined in the previous question - seems clear theoretically but maybe in this practical example there is a direct calculation to show it) $and$ that there exist $a,b,c,d$ s.t. $f^{\vee}$ can be written as $\left(\frac{ax +b}{cx+d}\right)$.
This last part should be doable using divisors - and maybe the RR theorem - but I'm at a loss as to how exactly. Perhaps it would be helpful to me to know what properties a function like $f^{\vee}$ must have on $\mathbb{P}^1$ that can be deduced by using divisors and the Riemann-Roch theorem. Maybe from those properties it is clear that it must be possible to write it that way.
As in the last question, you seem to keep jumping to divisors and RR. These are very useful and important things, but they're not necessary here either.
Let $f:\newcommand\PP{\mathbb{P}}\PP^1_k\to\PP^1_k$ be an automorphism. Fix homogeneous coordinates $(x:y)$ on $\Bbb{P}^1$. Then by construction of $f^\vee$, the zero set of $f^\vee$ is the set of points of $\Bbb{P}^1$ mapping to $(0:1)$ and the pole set of $f^\vee$ is the set of points of $\Bbb{P}^1$ mapping to $(1:0)$. Since $f$ is an automorphism, both of these sets are singletons. Since we know that the rational functions on $\Bbb{P}^1$ are fractions of homogeneous polynomials with numerator and denominator of the same degree. This immediately tells us that $f^\vee$ (expressed in lowest terms, using that $k[x,y]$ is a UFD) has both numerator and denominator of degree $1$. Hence $f^\vee = \frac{ax+by}{cx+dy}$. In the open affine with $y=1$, this is the rational function $\frac{ax+b}{cx+d}$.
If you regard the rational functions on $\PP^1_k$ as being the rational functions on the open affine with $y=1$, then the definition of $f^\vee$ is that it is the pullback of the regular function $x$ regarded as a rational function on all of $\PP^1$. Thus the induced map of function fields is indeed $x\mapsto \frac{ax+b}{cx+d}$.