i) A Weil divisor is a sum over the codimension 1 subvarieties, does that mean any point is generating the other points or are they all equally considered generators?
ii) This paper says that the sum's finite, https://www.math.ksu.edu/~ranno/math816/divisors.pdf I thought it would be infinite because the line has a compactification or something, how is it supposed to be finite again because it is in bijection with the natural numbers? Is the infinite point ignored in this?
The emphasis here is that you are taking formal linear combinations. As a reminder, given a set $\Sigma$ and a ground ring $A$, we define the free module $A-$module on $\Sigma$ to be $$A\langle \Sigma\rangle=\{a_1[\sigma_1]+\cdots a_k[\sigma_k]:\sigma_i\in \Sigma, a_i \in A\}.$$ Here $[\sigma]$ is a symbol corresponding to $\sigma \in \Sigma$. That is, we get one "basis" element for each element of $\Sigma$. The addition law is defined by $a[\sigma]+b[\tau]:=a[\sigma]+b[\tau]$ unless $\sigma=\tau$ in which case we get $(a+b)[\sigma]$. That is "combine like terms." When $A=\Bbb{Z}$, we call this the free Abelian group on $\Sigma$.
In the case of a variety $X/k$, we define $\operatorname{Div}(X)$ (Weil Divisors) to be the free Abelian group generated by the codimension $1$ subvarieties $Y_i\subseteq X$, so that $$ \operatorname{Div}(X)=\{\sum_{i=1}^k a_i[Y_i]:Y_i\subseteq X\:\text{is a codimension $1$ subvariety}\}.$$ In the case which I think you are asking about, we can work with $\Bbb{P}^1(\Bbb{C})=\Bbb{P}^1$, viewed as a compactification of $\Bbb{A}^1=\Bbb{C}$. Then, the codimension $1$ subvarieties are points. So, $$ \operatorname{Div}(\Bbb{P}^1)=\{\sum_{i=1}^ka_i[P_i]:P_i\in \Bbb{P}^1\}.$$ Note however that we cannot combine $[P_i]+[P_j]$ in any way besides the formal way stated above. There are no relations on the generators of this Abelian group. The sum here consists of finitely many terms by definition, so that in particular $$ [0]+3[1]+2[\infty]$$ is perfectly well-defined and finite, being a sum with $3$ terms.