I am referring to this question here:Prove Riemann-Roch theorem for Riemann sphere and torus
Where the OP would like to proof that $\dim H^0(\mathbb{P}^1, \mathcal{O}_D) = 1 + deg(D)$. I read through the answer posted by Y.Hu but I am unable to understand why he concluded that $f\in H^0(\mathbb{P}^1, \mathcal{O}_D)$ is of the form $f = z^{-k}h$. I feel like I am missing out something very obvious here. Can anyone help to explain that in greater detail?
Let $D = \sum_{j=1}^J n_j P_j$ a formal sum of points on the Riemann sphere $\Bbb{P}^1 = \Bbb{C} \cup \infty$, whose field of rational functions is $\Bbb{C}(z)$.
For open sets $U \subset \Bbb{P}^1$ let the sheaf $O_D(U) = \{ f \in \Bbb{C}(z), Div(f)|_U + D|_U \ge 0\}$ and its global sections $H^0(\Bbb{P}^1,O_D) = \bigcap_{U \subset \Bbb{P}^1} O_D(U)$.
For $h \in \Bbb{C}(z)$ we have that $f \mapsto fh$ is an isomorphism $O_D \to O_{D-Div(h)}$ and $H^0(\Bbb{P}^1,O_D) \to H^0(\Bbb{P}^1,O_{D-Div(h)})$. Since there is the rational function $z-P_j$ whose divisor is $P_j-[\infty]$, the isomorphism class of $O_D$ depends only on $N=\deg(D) = \sum_{j=1}^J n_j$.
Then $H^0(\Bbb{P}^1,O_{N [\infty]})$ is the set of polynomials $\in \Bbb{C}[z]$ of degree $\le N$ (if $N$ is negative this is empty)
and with $h = \prod_{j=1}^J (z-P_j)^{-n_j}$ then $H^0(\Bbb{P}^1,O_D) = h \ \cdot \ H^0(\Bbb{P}^1,O_{N [\infty]})\}$.
To see how it relates to Riemann Roch you need the divisor of $df$ for some rational function, for example $K=Div(dz) = -2[\infty]$, and to know the genus is $g=0$.