I understand what a discrete valuation and a normalized discrete valuation are. I know that for an algebraically closed field $k$, the projective line $\mathbb{P}^1_k = \text{Proj}(k[x,y])$ is defined as the set of homogeneous prime ideals of $k[x,y]$ with similar topology and such to $\mathbb{A}_k^1$, etc. Here's the statement of the problem in Görtz and Wedhorn's Algebraic Geometry:
Let $k$ be an algebraically closed field and let $X = \mathbb{P}_k^1$ be the projective line over $k$. Its function field is $k(T)$, the field of fractions of the polynomial ring $k[T]$.
For each monic irreducible polynomial $p\in k[T]$ let $v_p$ be the $p$-adic valuation on $k(T)$. Define also the discrete valuation $v_{\infty}$ on $k(T)$ by $v_{\infty}(f/g) = \text{deg}(g) - \text{deg}(f)$, if $f,g\in k[T]$ are nonzero polynomials. Show that these valuations are the normalized discrete valuations corresponding to the closed points $X_0$ of $X$.
Closed points of $\text{Proj}(k[x,y])$ for $k$ algebraically closed are just points $[(ax - by)]$ where at least one of $a$ and $b$ is non-zero. So for $a\neq 0$ we just dehomogenize with respect to $y$ and have $T = x/y$ so that the discrete valuation $v_{[(ax - by)]}$ just takes an element $f/g\in K = k(T)$ to the integer $n$ where $(T - b/a)^nf'/g' = f/g$ and $f',g'\not\in (T - b/a)$. I think the valuation $v_{\infty}$ is supposed to correspond to the point at infinity, meaning the point $[(y)]$ (like $[0:1]$ in the classical setting). But... I don't know how to show that at all, much less how to get the degree popping up out of the valuation corresponding to $[(y)]$.
My question is: Should $v_{\infty}$ correspond to the closed point $[(y)]$? Any hints as to how I can get to a discrete valuation like $v_{\infty}$ from the closed point $[(y)]$ (i.e. why the answer to the first question haha)?
In your definition of the discrete valuation, you have been using $T - c$ as the uniformizing parameter for the local ring of the point $P = [a:b]$ where $T = x/y$ and $c = a/b$. Unfortunately, this doesn't work at the point $[1:0]$ since $T$ has a pole there. Instead, we can take the function $S := 1/T = y/x$ as our uniformizing parameter. Note that $k(T) = k(1/T) = k(S)$. Thus to determine the order of vanishing of a function $f/g \in k(T)$ at $[1:0]$, it suffices to rewrite the function in terms of $S$.
Given $f = \sum_{i=0}^n a_i T^i \in k[T]$ with $\deg(f) = n$, then \begin{align*} f(T) &= a_0 + a_1 T + \cdots + a_n T^n = a_0 + a_1 (1/S) + \cdots + a_n (1/S)^n = \frac{a_0 S^n + a_1 S^{n-1} + \cdots + a_n}{S^n}\\ &= S^{-n} (a_0 S^n + a_1 S^{n-1} + \cdots + a_n) \, . \end{align*} Given $g = \sum_{i=0}^m b_i T^i \in k[T]$ with $\deg(g) = m$, then for the rational function $f/g$ we have \begin{align*} \frac{f(T)}{g(T)} &= \frac{S^{-n} (a_0 S^n + a_1 S^{n-1} + \cdots + a_n)}{S^{-m} (b_0 S^m + b_1 S^{m-1} + \cdots + b_m)} = S^{m-n} \frac{a_0 S^n + a_1 S^{n-1} + \cdots + a_n}{b_0 S^m + b_1 S^{m-1} + \cdots + b_m}\\ &= S^{\deg(g) - \deg(f)} \frac{\widetilde{f}}{\widetilde{g}} \end{align*} and just as in the definition you gave above, $S$ doesn't divide $\widetilde{f}$ or $\widetilde{g}$.