Let's analyze the proof of the following lemma:
Lemma: Let $X$ be an integral, Noetherian scheme and let $f\in K(X)^\ast$, then for all but finitely many points $x\in X$ of codimension $1$ we have that $f_x\in\mathcal O_{X,x}^\ast$.
"Proof": We know that $K(X)=\text{frac }\mathcal O_X(U)$ for any open set $U\subset X$. So let's fix an affine open $U$, then $f=\frac{a}{b}$ for $a,b\in\mathcal O_X(U)$.
(first unclear thing): for a point $x$ of codimension $1$, if $$x\notin V(a)\cup V(b)\cup (X\setminus U)\quad (*)$$ then $f_x\in\mathcal O_{X,x}^\ast$. I don't understand the implication, could you please make it clearer?
(Second unclear thing): In some way one shows that the points not satisfying $(\ast)$ are finitely many. Could you please explain this fact? (I suppose that here the Noetherian property of $X$ is used )
Let $R = \mathcal{O}_X(U)$.
Take $x\in U$, with $\mathfrak{p} \subset R$ the corresponding prime ideal. If $x\notin V(b)$, this says that $b\notin \mathfrak{p}$, so $b_\mathfrak{p}$ is a unit in $R_\mathfrak{p}$.
Likewise, if $x\notin V(a)$, then $a_\mathfrak{p}$ is a unit in $R_\mathfrak{p}$. So $a_\mathfrak{p} / b_\mathfrak{p}$ is a well-defined unit in $R_\mathfrak{p}$. Since we can identify $R_\mathfrak{p}$ with the subring $\mathcal{O}_{X,x}$ of $K(X)$, 1) follows.
For 2), we can use the fact that, in a noetherian ring, there are only finitely many minimal primes over a given ideal. For example, this follows from primary decomposition.