I think your first example for $n=2$ fails. Conics on the quadric surface are exactly the plane sections, but if $p$ and $q$ are on the same ruling of the quadric surface, then any plane containing both of them also contains the line of the ruling that joins them. So there is no smooth conic containing both. Am I making sense?

It is definitely asking too much for this to hold for all $t \in C$. Even if $f$ is smooth, this will typically only hold for $t$ near 0.

From stacks.math.columbia.edu: "The Stacks Project now consists of:...5734 pages..." That may not be what the OP had in mind when using the word "precise".

That's right: if you know the fibres have the same Hilbert polynomial, then the family is flat. I was suggesting that one could deduce constancy of the Hilbert polynomial by showing (in another way) that the family is flat, but Lvovski's argument seems simpler.

I don't understand how "then it works" can be an answer to "what can we say about $X$?" Can you clarify exactly what the question is?

I believe so, but there is something to check, namely that the total space of the family over that open subset is not too singular. I don't see how to do that right now, so my argument currently gives a weaker result than Lvovski's.

Inside $G(r,h^0) \times X$ you can form the family whose fibre over an $r$-dimensional subspace $V$ is the closed subscheme $Z_V$. By generic flatness there will be a dense open subset in $G(r,h^0)$ over which this is flat, hence any two subschemes which are fibres over points in the open subset will have the same Hilbert polynomial.

To make Sergey's answer even more concrete, try an example such as the Fermat quartic in $\mathbf P^3$. Here an elliptic fibration is given by projecting away from a line on the surface; any other line that is disjoint from the projection centre will give a section. But you can easily find two such lines that are not disjoint from each other.

If the central fibre is singular it definitely need not be homeomorphism (I assume we are talking about the analytic topoology). For example, a family of smooth conics degenerating to a pair of lines.

@YCor: do you also talk about Ligne--Mumford stacks? :)

@JasonStarr: I would like to say that Hacon-McKernan's proof of Shokurov's conjecture implies that if $X$ is log terminal, then $CH_0 \tilde{X}$ is still isomorphic to $CH_0 X$. (So your counterexample is somehow as good as one can do.) Maybe I am missing a subtlety.

The accepted answer at mathoverflow.net/questions/241860/… proves birational invariance of $CH_0$ for smooth proper varieties (over any field).

OK, thanks for the extra infomation!

Is this 12-point configuration just the dual of the Hesse configuration?

I like the first one. It seems pretty standard: for example the Wikipedia page en.wikipedia.org/wiki/Contact_(mathematics) uses more or less this formulation.