Now your actual question seems to be:
As Hartshorne chapter III.9.2 claim，an Ox-module （need not be quasi coherent) F's flatness is stable under base change. But the stalks is not the tensor products, how can I prove the claim?
The statement is the following: If $f : X \to Y, Y' \to Y$ are morphisms, and $\mathcal{F}$ is a module over $X$ which is flat over $f$, then the pullback of $\mathcal{F}$ to $X \times_Y Y'$ is flat over $X \times_Y Y' \to Y'$. I am pretty sure that Hartshorne understands $\mathcal{F}$ to be quasi-coherent here. Otherwise the sketch of proof also does not make sense. But it is also true in general:
Pick a point in $X \times_Y Y'$, thus a triple $(x,y',\mathfrak{p})$ as described above. Let $y$ be the underlying point in $Y$. Now $\mathcal{F}_{x}$ is flat over $\mathcal{O}_{Y,y}$. By commutative algebra (base change of flat modules), it follows that $\mathcal{F}_x \otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{Y',y'}$ is flat over $\mathcal{O}_{Y',y'}$. Again by commutative algebra (localizations are flat) $(\mathcal{F}_x \otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{Y',y'})_{\mathfrak{p}}$ is flat over $\mathcal{O}_{Y',y'}$. But this is exactly the stalk of the pullback of $\mathcal{F}$ in the given point $(x,y',\mathfrak{p})$.
