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Charles, in your answer you're basically discovering the fact that the normalization is not flat (answer edited to show that it actually does provide an answer to the original question) Let $X$ be a n …

Take the simple flatness criterion that Theorem Let $f:X\to B$ be a morphism such that $B$ is integral, regular, and of dimension $1$. … Most other flatness criteria require something about the fibers of the map, which is again only topological on the base, but more than that on the total space and/or fibers. …

I think there is some confusion here. Either on your part or on mine. I don't think being non-reduced is equivalent to having a non-reduced component. A scheme may have a fat point, but be irreducible …

You need to assume also that $f$ is non-isotrivial. (Think of $X=S\times F$). On the other hand, you don't need the semi-stable assumption for any of these, except that the explicit bound you are aski …

This is false in general. Take $f:\mathbb A^2\to \mathbb A^1$, $(x,y)\mapsto xy$. Any attempt to resolve the singular fiber will bring in a new component in the fiber, so it remains singular.

Let's say that $X={\rm Spec\,} k$ for a field $k$. Then it is certainly Gorenstein and $k$ is an injective sheaf on $X$. For any $S$ and $p$ as defined in the question, $p^*k\simeq \mathscr O_S$. If t …

You already have trivial counterexamples for your statement, but perhaps you were thinking of a section whose zero locus is irreducible and dominates $Y$. It is false even with that additional assumpt …