Conditions for faithful flatness I was looking for a proof of the folklore fact that given a flat ring homomorphism $R \rightarrow S$ then $S$ is faithfully flat iff the induced map on the spectra is surjective. I found one on the Stacks project but there are some points I do not understand fully. 
https://stacks.math.columbia.edu/tag/00HP
My first doubt is why $M \otimes \kappa(\mathfrak{m})=M/\mathfrak{m}M$. This is not true for general $M$: if $M$ had $(R \setminus \mathfrak{m})$-torsion then the first term would be zero. Being $M$ flat it is torsion-free thus this cannot be. I can see that the quotient by $\mathfrak{m}$ gives the quotient in the module but I really do not understand why we can ignore the localization.
My second question is about the idea behind the proof: the slogan is "a flat module is faithful iff it has no zero fibers". The idea of thinking of the tensor with the residue fields $\kappa(\mathfrak{p})$ as a fibre comes from the following observation
https://stacks.math.columbia.edu/tag/00E6
Where it is proved that given a map $f \colon R \rightarrow S$ then $\mathfrak{p}$ is in the image of $Spec(f)$ iff $S \otimes \kappa(\mathfrak{p}) \neq 0$. The argument relies on the fact that the lower squares induce pull-back squares on the spectra of the involved rings, but if the ideal $\mathfrak{p}$ has no preimage in $Spec(S)$ then $S/\mathfrak{p}$ and $S_\mathfrak{p}$ are a priori $R$-modules. For $S/\mathfrak{p}$  I think we can interpret it as the quotient of $S$ by the ideal generated by $\mathfrak{p}$ so it has a ring structure.
But for $S_\mathfrak{p}$ I am unsure: usually we localize by inverting the elements of a multiplicatively closed subset of the ring and I cannot see what this set is here. $S \setminus f(\mathfrak{p} )$ or $f(R) \setminus f(\mathfrak{p} )$?
I also have 2 other questions related to the lemma, but not to its proof.
Do you have in mind any counterexamples for the equivalence of conditions 2 and 3 of lemma if we drop the flatness assumption? That is, can you provide an example of a non-flat ring homomorphism whose induced map between spectra is surjective on the closed points but not on all the elements of $Spec(R)$?
What is the geometric meaning of the flatness condition on the map $f$? We saw that faithfully flatness is equivalent to $Spec(f)$ being surjective so we can consider $Spec(S)$ as a cover of $Spec(R)$. Do we any interesting situation on the affine schemes if $f$ is just flat?
 A: First, the statement $M\otimes_R \kappa(\mathfrak{m})=M/\mathfrak{m}M$ is true for any maximal ideal $\mathfrak{m}$. This is because $\kappa(\mathfrak{m})=R/\mathfrak{m}$, and $M\otimes_R R/I=M/IM$ is true for any ring $R$, ideal $I\subset R$, and $R$-module $M$. One may ignore the localization because $R/\mathfrak{m}\cong R_\mathfrak{m}/\mathfrak{m}_\mathfrak{m}$ for maximal ideals $\mathfrak{m}$ (exercise: prove this).
For the second question, we localize by $f(R\setminus\mathfrak{p})$, as one can see on the right hand side of the diagram.
As for the question of a ring homomorphism which is surjective on maximal ideals but not surjective, sure: for any DVR $R$ with maximal ideal $\mathfrak{m}$, the map which is $\operatorname{Spec}$ of $R\to R/\mathfrak{m}$ is surjective on closed points but not surjective: $\operatorname{Spec} R$ has two points, one of which is closed, and the other of which is not. The map hits the closed point, but not the generic point.
Finally, the geometric meaning of flatness can be better explained by the following MO post, so I'm just going to link it: Why are flat morphisms flat?
