$\mathcal{O}_{X,\eta}$ is a regular local $1$-dimensional noetherian domain. It is a Theorem in commutative algebra which says that this is precisely a DVR.
If $X$ is an arbitrary integral scheme and $x \in X$, then the quotient field of $\mathcal{O}_{X,x}$ is the function field of $X$. Namely, since this a local issue, we may assume $X=\mathrm{Spec}(A)$ for some integral domain $A$, and just have to observe that $\mathrm{Quot}(A_{\mathfrak{p}}) = \mathrm{Quot}(A)$ for every prime $\mathfrak{p} \subseteq A$.
As for the last question, you should look at the definitions. Nothing happens.
Let's understand the subsequent arguments of the proof in steps. I will use the notation in your link's answer (, in which $T_x = \mathrm{Spec}\mathcal{O}_x$, and $X^{(1)}$ is the set of 1-codimensional points of $X$).
- The principal divisor $(f_x)$ on $X$ has the same restriction to $\mathrm{Spec}\mathcal{O}_x$ as $D$, hence they differ only at prime divisors which do not pass through $x$.
Recall that $D_x=\sum_{y\in X^{(1)}\cap T_x}n_y\cdot(\overline{\{y\}} \cap T_x)$, and that as a principal divisor in the $X$, $(f_x)=\sum_{y\in X^{(1)}}v_y(f_x)\cdot \overline{\{y\}}$, where $v_y$ is the valuation on the prime divisor $\overline{\{y\}}$.
Then note that the valuation of $\overline{\{y\}}$ is just the valuation of $\overline{\{y\}}\cap T_x$, since they have the same local rings on $y$. Thus the restrictions of $D_x$ and $(f_x)$ are both equal to $\sum_{y\in X^{(1)}\cap T_x}v_y(f_x)\cdot(\overline{\{y\}}\cap T_x)$, and globally they differ just at those points not in the $T_x$. But it's clear that a prime divisor doesn't pass through $x$, if and only if its generic point is not in the $T_x$.
- There are only finitely many of these which have a non-zero coefficient in $D$ or $(f_x)$, so there is an open neighbourhood $U_x$ of $x$ such that $D$ and $(f_x)$ have the same restriction to $U_x$.
That's really clear because $T_x$ is just the intersection of all open neighbourhoods of $x$, and replacing $T_x$ by a neighbourhood $U_x$ won't change the coefficients in both divisors.
- Covering $X$ with such open sets $U_x$, the functions $f_x$ give a Cartier divisor on $X$. Note that if $f$,$f'$ give the same Weil divisor on an open set $U$, then $f/f'\in\Gamma(U,\mathcal{O}^*)$, since $X$ is normal.
The second sentence needs to be verified. $(f)=(f')$ implies that $(f/f')=0$, whence in each $y\in X^{(1)}$, $f/f'$ is a unit of $\mathcal{O}_y$. For every affine open subset of $X$, say $\mathrm{Spec}A$, I claim that $f/f'$ is a global section on it, i.e. $f/f'\in A$. Indeed, each prime ideal $\mathfrak{p}$ of $A$, of height $1$, corresponds to a point of codimension $1$ in $X$, and thus $f/f'\in A_\mathfrak{p}$. Since $A$ is a normal noetherian domain, by commutative-algebra theories, we have
$$A=\bigcap_{\mathrm{ht}\mathfrak{p}=1}A_\mathfrak{p}.$$
Therefore $f/f'\in A$. Then given an affine open covering $\{U_i\}$ of $U$, $(f/f')|_{U_i}\in\Gamma(U_i,\mathcal{O})$, so one can glue them into a section in $\Gamma(U,\mathcal{O})$. Actually this section is just $f/f'$ because they must be the same in $\mathscr{K}$. Thus $f/f'\in\Gamma(U,\mathcal{O})$. Similarly, $f'/f$ is also in the $\Gamma(U,\mathcal{O})$. So we have $f/f'\in\Gamma(U,\mathcal{O}^*)$.
Best Answer
It's very much possible to describe this more explicitly. I will quote Eisenbud from his book Commutative Algebra with a view toward algebraic geometry for a precise statement:
Think of this as follows: Your prime divisor $Z$ is the closure of a point $P$ of codimension one. If you go into a local neighbourhood of it, there is a single function $\pi$ that defines its vanishing. More precisely, think of $X=\mathrm{Spec}(A)$. If you take any rational function $f=a/b$ with $a,b\in A$, then the stalks of $a$ and $b$ at $P$ both have a well-defined order of vanishing, in the sense that $a=u\pi^r$ and $b=v\pi^s$ with $u$ and $v$ units in $\mathcal O_{X,P}$. Then, $a$ vanishes at $P$ to the order $r$ and $b$ to the order $s$. If $r>s$, then $f$ will vanish at $P$ to the order $r-s$. If $s>r$, then $f$ has a pole at $P$, of order $s-r$.
Silly example: Consider the rational function $$ f=\frac{(x-4)(y-2)^3(x+6)}{(x-4)^5(y-2)} $$ on $\mathbb A^2$. Then $f$ has order $-4$ along the line $x=4$ and it has order $1$ along the line $y=2$. The example is silly because the message is the following: You are usually not able to write your prime divisor as the vanishing of one single global function, and even if you could, your ring might not be factorial, etc etc. The bottom line is that everything works very nice when you go into a neighbourhood of your point that is local enough, and there it is the same as back in highschool, just count how often the zero appears in the numerator and how often it appears in the denominator, subtract, and then you know if it's a zero or a pole. And you can even say how much of either it is.