Below is a proposition from here.
The proof goes like this:
First, I can't see how $(1)\implies(2)\iff(3)$ is true. Certainly $0\ne F/PF=F\otimes_R R/P$, but the RHS of the latter is supposed to be equal to $R_P/PR_P\otimes_R F$. However, $R_P/PR_P\otimes_R F=(R/P)_P\otimes_R F=(R/P\otimes_R F)_P=R_P/{PR_P}\otimes_{R_P} F_P$. So I can't see why $F\otimes_R R/P=R_P/PR_P\otimes_R F$ holds.
Second, how is the middle arrow in the last sequence of the proof defined? (I can't see why this sequence is exact.) Then I believe the fact that $M\otimes_R F\ne 0$ is inferred from the fact that the middle arrow is an isomorphism and its domain is not the zero module, isn't it? If so, I wonder why the domain is not the zero module?
Finally, will this proposition hold if "maximal" is replaced with "prime"?
There is no localization involved.
Suppose (1) holds. Then $R/P\ne0$ implies $(R/P)\otimes F\ne0$; but $(R/P)\otimes F\cong F/PF$, so $PF\ne F$. Hence we have (2).
Suppose (2) holds. If $M\ne0$ and $x\in M$, $x\ne0$, consider a maximal ideal $P$ such that $I=\operatorname{Ann}(x)\subseteq P$. Then the surjection $R/I\to R/P$ yields a surjection $(R/I)\otimes F\to (R/P)\otimes F$. Therefore $IF\ne F$. It follows that $xR\otimes F\ne0$, since $xR\cong R/I$. By flatness, $xR\otimes F$ embeds in $M\otimes F$, so also $M\otimes F\ne0$. Hence (1) holds.
(2)$\iff$(3) is done similarly.
You can substitute “maximal ideal” with “proper ideal” for (1)$\iff$(2). Of course, (3) doesn't make sense with $P$ non prime, but so long as $k(P)\cong R/P$, the two conditions (2) and (3) are clearly equivalent.