$\mathcal{O}_{X,\eta}$ 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.
First, what is a nonzerodivisor, geometrically? Say we start with a variety $X$, possibly with many irreducible and/or embedded components $X_i$. "Modding out by a nonzerodivisor" means cutting $X$ with a hypersurface in so that the following is true:
$$\dim (X_i \cap H) < \dim X_i \text{ for every associated component } X_i \subseteq X.$$ (Algebraically: an element $f \in R$ is a nonzerodivisor iff $f \notin P$ for all associated primes $P$, iff $\dim R/(P + (f)) < \dim R$ for all associated primes $P$.)
Then, the geometric meaning of regular sequences is the following: we try to repeat this process with a sequence of hypersurfaces. That is, at each step, we require that $H_i$ cut down the dimension of every associated component of $X \cap H_1 \cap \cdots \cap H_{i-1}$. This is much stricter than, say, just lowering the dimension of $X$ by 1 at each step (in fact that latter leads to the weaker notion of system of parameters.)
The reason this is useful to study locally is that regular sequences of length equal to the ring's dimension are generalizations of 'local coordinates', allowing for mild singular points. In fact Cohen-Macaulay rings are sometimes described as "algebraically smooth" or "homologically smooth", since they share many homological properties with regular local rings.
So, we pick a point $p \in X$, and ask that our hypersurfaces $H_i$ cut out $p$ in the manner described above. Now there is a problem: if $p$ itself becomes an embedded point partway through, then we certainly won't be able to 'cut $p$ down by a dimension', so we'll be stuck. If $X$ is not equidimensional at $p$, or if $X$ has any embedded components, then this is bound to happen (the lowest-dimensional component will reach dimension 0 before the largest one does, at which point $p$ will become an embedded point). And this is really the only obstacle. In fact, we could have defined Cohen-Macaulay recursively in this way: all zero-dimensional rings are Cohen-Macaulay, and for the others we have:
Definition. A positive-dimensional local ring $R$ is Cohen-Macaulay if (a) it is equidimensional, with no embedded associated primes, and (b) for some (equivalently, any) nonunit nonzerodivisor $f \in R$, the ring $R/f$ is Cohen-Macaulay.
In response to your questions (1) and (2), local complete intersections are slightly nicer than Cohen-Macaulay rings, since they are of the form [regular local ring] / [regular sequence]. (Homological algebra tells us that we can extend this regular sequence to one of maximal length, which is why LCIs are Cohen-Macaulay.) So I guess LCI is about cutting a variety $X$ out of a smooth ambient space by regular sequence, while Cohen-Macaulay is about being able to cut $X$ down to a point by a regular sequence.
Finally, in your example of $R = k[x,y,z]/(xz,yz)$, the regular sequence $xy+3, x-y$ cuts out the points $(i\sqrt{3},i\sqrt{3},0)$ and $(-i\sqrt{3},-i\sqrt{3},0)$. So we now know that $R$ is Cohen-Macaulay at those two points. This doesn't tell us what's going on at other points, like the origin, since the hypersurfaces just miss those points. (And in fact $R$ is Cohen-Macaulay at every point except the origin.)
Best Answer
I believe what is going on is point-set topology, because of the following:
Lemma. Let $X$ be a Jacobson space and consider a locally quasi-constructible subset $U \subseteq X$. If $Y \subseteq X$ is a locally quasi-constructible subspace such that $U \cap Y \ne \emptyset$, then there exists a point $x \in Y$ that is closed in $X$ such that $x \in U$.
Proof. The set $U \cap Y$ is locally quasi-constructible in $Y$ by definition of the subspace topology, and is non-empty by hypothesis. Since $Y$ is a Jacobson topological space [EGAI$_{\text{new}}$, Chapitre 0, Proposition 2.8.2], we see that $U \cap Y$ contains a closed point $x \in Y$ by definition of a Jacobson space [EGAI$_{\text{new}}$, Chapitre 0, Définition 2.8.1]. This point $x$ is closed in $X$ by [EGAI$_{\text{new}}$, Chapitre 0, Proposition 2.8.2]. $\blacksquare$
To apply this to your situation, every scheme of finite type over a field $k$ is Jacobson [EGAI$_{\text{new}}$, Proposition 6.5.2], and setting $U$ to be the regular locus $\operatorname{Reg}(X)$, we see that $\operatorname{Reg}(X)$ is open, hence locally quasi-constructible.
Of course, maybe it is easier to just prove your statement directly: The set $\operatorname{Reg}(X) \cap Y$ is open in $Y$ by definition of the subspace topology, and is non-empty since it contains $\eta$. Since closed points are dense in $Y$ [EGAI$_{\text{new}}$, Proposition 6.5.2], there exists a point $x \in \operatorname{Reg}(X) \cap Y$ that is closed in $Y$. Finally, this point $x$ is closed in $X$ since $\{x\} \hookrightarrow Y \hookrightarrow X$ is a composition of closed embeddings.