In general, separated morphisms are stable under composition and under base change. This follows from the definitions, no valuative criterion is necessary here. You can also find this in every detailed introduction to schemes. This implies (c). Open immersions are seperated (in fact, every monomorphism is separated). This implies (a). Affine morphisms are separated. This gives the one direction of (b). Now assume that $X$ is separated over $R$, i.e. $X \to X \times_R X$ is a closed immersion. Then for every pair of open affines $U,V \subseteq X$ we have that $U \times_R V$ is affine, hence the preimage $U \cap V$ is affine and $\Gamma(U \times_R V) = \Gamma(U) \otimes_R \Gamma(V) \to \Gamma(U \cap V)$ is surjective. But this implies that also $\Gamma(U) \otimes_\mathbb{Z} \Gamma(V) \to \Gamma(U \cap V)$ is surjective. From this we see that $X$ is separated over $\mathbb{Z}$. Actually a more general statement is true: If $f \circ g$ is separated, then $g$ is separated (the standard proof factors $g$ over its graph).
I am not completely sure that I understand what it is you are looking for, but let me give an answer in the hope that you find it helpful.
Remember that base change is constructed (e.g. in Hartshorne) first in the case of morphisms of affine schemes, and the general case then follows by a rather involved glueing process. So to avoid the complications, I will modify your example to the case of affine schemes instead, by taking the open subscheme $\{ Z \neq 0 \}$ inside your family of curves.
Let me also change the names of your fields so that $k$ is the field we start with, and $K=k(a,b)$.
We then have a family $$ X \subset \mathbf A^4_k$$ cut out by the equation $$ax^2+by=0;$$
in other words $$X = \operatorname{Spec} \left( \frac{k[x,y,a,b]}{I} \right) $$
where $I$ is the ideal generated by $ax^2+by$.
Now the projection morphism $$\pi : X \rightarrow \mathbf A^2 \\ (x,y,a,b) \mapsto (a,b)$$
corresponds to the map of rings
$$ k[a,b] \rightarrow \frac{k[x,y,a,b]}{I}.$$
On the other hand, the inclusion of the generic point
$$ \iota: \operatorname{Spec} K \rightarrow \mathbf A^2_k$$
corresponds to the ring map
$$ k[a,b] \hookrightarrow K$$
embedding the polynomial ring in its field of fractions.
Finally, the crux of the matter: the base change of $\pi$ along $\iota$ is (by definition) given by the tensor product
$$ K \otimes_{k[a,b]} \frac{k[x,y,a,b]}{I}$$
which is isomorphic to
$$\frac{K[x,y]}{J}$$
where now $J$ is the ideal in $K[x,y]$ generated by the polynomial $ax^2+by$.
So the generic fibre is indeed defined by the same equation, but now viewed as a polynomial over a different field.
I hope that helps.
Best Answer
As pointed out in the comments, things can get a little strange when $X$ is reducible or $f$ isn't dominant. We'll treat the case where $X$ is irreducible and $f$ is dominant while trusting the reader to make the correct generalization of this to the case of $X$ reducible (treat each component $X_{\alpha}$ individually via the composite of the closed immersion $X_\alpha\to X$ and $f$, etc).
Here's a few results we'll need:
In addition to the hypotheses in the problem statement, we assume $X$ irreducible, $f$ dominant, and $W$ a proper closed subscheme.
By two applications of generic flatness (to $\mathcal{F}=\mathcal{O}_X$ and $\mathcal{F}=\mathcal{O}_W$), we may assume that both $f:X\to Y$ and $f|_W:W\to Y$ are flat by replacing $Y$ with a slightly smaller open subscheme which doesn't affect our calculations at the generic point*. Let $n=\dim X$ and $m=\dim Y$. If there exists a $w\in W$ so that $\dim_w W_{f(w)} \geq n-m$, then we have $\dim_w W \geq n-m+m = n$, which contradicts the fact that $W$ is a proper closed subscheme. So for all $w\in W$, $\dim_w W_{f(w)} < n-m$, so in particular $\dim W_\eta < n-m = \dim X_\eta$ and we're done.
If you're looking for more intuition about what should happen with respect to fibers of a morphism, Stacks has a good section on it here.
*: If this makes $W=\emptyset$, we're already done and we can stop right here - h/t Ben in the comments pointing out I didn't write this down.