Let $R\to S$ be a flat ring extension. By theorem 9.5 of the book Commutative Ring Theory written by Matsumura the going-down theorem holds between $R$ and $S$.
Is it true (or not) about these theorems: "INCOMPARABILITY THEOREM", "LYING-OVER THEOREM" and "GOING-UP THEOREM"? How about faithfully flat ring extensions?
On
Because the counterexamples to Incomparability and Going-Up in Alex Youcis' answer have positive-dimensional fibers, and because Incomparability is rescued by the assumption that the ring map $R\rightarrow S$ is of relative dimension zero, readers may wonder (as I did for a bit) whether Going-Up is also rescued by this assumption. It is not:
Take $R = k[x]$ (with $k$ any field), and $S = k[x]\times k[x,x^{-1}]$. The ring map $R\rightarrow S$ will be given by the composition
$$ k[x]\xrightarrow{\Delta} k[x]\times k[x] \xrightarrow{\pi} k[x]\times k[x,x^{-1}],$$
where $\Delta$ is the diagonal embedding $\Delta(f) = (f,f)$, and $\pi$ is the identity on the first factor and the canonical inclusion $k[x]\hookrightarrow k[x,x^{-1}]$ on the second. Now $\Delta$ is a flat map because $k[x]\times k[x]$ is free as a module over the image of $\Delta$, with basis $(1,0)$ and $(0,1)$. (Indeed, arbitrary $(f,g)\in k[x]\times k[x]$ is uniquely represented as $\Delta(f)(1,0) + \Delta(g)(0,1) = (f,f)(1,0) + (g,g)(0,1)$.) And $\pi$ is also a flat map because it is the direct product of the identity and the flat map $k[x]\rightarrow k[x,x^{-1}]$. Therefore the composition $\pi\circ\Delta: R\rightarrow S$ is flat.
It is even faithfully flat because the induced map on $\operatorname{Spec}$s is surjective, i.e., it satisfies Lying-Over. This can be seen because for any prime $\mathfrak{p}$ of $R=k[x]$, the prime ideal $\mathfrak{p}\times k[x,x^{-1}]$ in $S$ pulls back to $\mathfrak{p}$ in $R$.
Furthermore, it is of relative dimension zero. This can be seen completely explicitly. If $\mathfrak{p}$ is any prime ideal of $k[x]$ not containing $x$ (i.e. any prime ideal except $(x)$), then the fiber $S\otimes_R \kappa(\mathfrak{p})$, where $\kappa(\mathfrak{p})$ is the residue field at $\mathfrak{p}$, is the ring $\kappa(\mathfrak{p})\times \kappa(\mathfrak{p})$, which is zero-dimensional because it is a product of fields. If $\mathfrak{p}$ contains $x$ i.e. equals $(x)$, then $S\otimes_R \kappa(\mathfrak{p})$ is $\kappa(\mathfrak{p})\times 0 = k\times 0 \cong k$, again a zero-dimensional ring.
(It might be easier to understand it geometrically. $R$ is the coordinate ring of a line; $S$ is the coordinate ring of two disjoint lines, one of which is missing a point; and the map $R\rightarrow S$ is induced from projecting the two lines down to the one. If $k$ is algebraically closed, then the elements of the Specs are exactly the points of these lines, plus a generic point for each line. The fiber over any point of the downstairs line is a pair of points of the upstairs lines, unless the point downstairs is under the hole upstairs, in which case the fiber is a single point. In either case, the fiber is zero-dimensional. The situation with the generic points, which is the case $\mathfrak{p}=0$, so $\kappa(\mathfrak{p})=k(x)$, and $S\otimes_R \kappa(\mathfrak{p}) = k(x)\times k(x)$, just captures the generic situation with the closed points.)
Anyway, although $R\rightarrow S$ is faithfully flat and of relative dimension zero, it does not satisfy going-up. In particular, the prime ideal $k[x]\times 0$ in $S$ lies over the zero ideal in $R$, but there is no prime ideal of $S$ that contains this one and lies over $(x)$ in $R$, because it would have to contain both $\Delta(x) = (x,x)$ and also $(1,0)$, but then it would contain $(x,x) - (x,0)(1,0) = (0,x)$, and then it would $(0,x)(0,x^{-1})=(0,1)$, and then it would contain $(1,0)+(0,1)=(1,1)$, which is the multiplicative identity in $S$. (Geometrically, $0\subset R$ is the generic point and $(x)\subset R$ is the point $x=0$, and the inclusion $0\subset (x)$ in $R$ witnesses the fact that we can specialize the generic point to $x=0$. The generic point of the line-missing-a-point from $\operatorname{Spec}S$ projects to the generic point of $\operatorname{Spec}R$, but it cannot be specialized to a point that lies over $0$, because this point is missing.)
The lying over theorem holds for faithfully flat ring extensions. In fact, a flat ring map $R\to S$ is faithfully flat, if and only if $\text{Spec}(S)\to\text{Spec}(R)$ is surjective. So, a flat but unfaithful ring extension will not satisfy lying over.
For the incomparability, this is false on both counts. Consider $k\to k[x]$.
The going up theorem doesn't hold for faithfully flat ring extensions. For example, consider $\mathbb{Z}\to\mathbb{Z}[T]$. This is faithfully flat, since $\mathbb{Z}[T]$ is a free $\mathbb{Z}$-module. Consider then, for a prime $p$, the ideal $(1+pT)\subseteq\mathbb{Z}[T]$ sitting above $(0)$ in $\mathbb{Z}$. We can extend $(0)$ to $(p)$, but we can't extend $(1+pT)$ by $Q$ lying above $p$, else $Q$ would contain $1+pT$, and $pT$ and so would be $\mathbb{Z}[T]$.