A flat ring map is faithfully flat exactly when it detects proper ideals and is surjective on spectra

By [L2], the flatness of $f$ is the flatness of the $R$-module $S$. For a proper ideal $I$, the quotient $R\mathrm{/}I$ is nonzero, and $(R\mathrm{/}I){\otimes }_{R}S\cong S\mathrm{/}IS\mathrm{.}$ Therefore [L1] says that $f$ is faithfully flat exactly when $S\mathrm{/}IS\mathrm{\ne }0$ for every proper ideal $I$, which is equivalent to $IS\mathrm{\ne }S$. So 1 and 2 are equivalent.

A prime of $S$ lying over a maximal ideal $\mathfrak{m}$ exists exactly when the fibre ring $S{\otimes }_R\kappa (\mathfrak{m})$ is nonzero. Since $R\mathrm{/}\mathfrak{m}{\otimes }_{R}S\cong S\mathrm{/}\mathfrak{m}S$ and $\mathfrak{m}$ is maximal, this is precisely the maximal-ideal fibre condition in [L1]. Thus 1 and 3 are equivalent.

Assume 1 and let $\mathfrak{p}\subset R$ be prime. By the prime-fibre form of [L1], the ring $T:=S{\otimes }_R\kappa (\mathfrak{p})$ is nonzero. Choose a prime ideal $\mathfrak{n}\subset T$, and let $\mathfrak{q}$ be its preimage in $S$. Because the composite $R\to S\to T$ factors through $\kappa (\mathfrak{p})$, every element of $\mathfrak{p}$ maps to $0$ in $T$, so $\mathfrak{p}\subseteq \mathfrak{q}\cap R$. If $r\in R\setminus \mathfrak{p}$, then its image in the field $\kappa (\mathfrak{p})$ is a unit, hence its image in $T$ is a unit and cannot lie in the prime $\mathfrak{n}$. Therefore $r\mathrm{\notin }\mathfrak{q}$, so $\mathfrak{q}\cap R=\mathfrak{p}$. Thus a prime of $S$ lies over $\mathfrak{p}$, and 1 implies 4.

Condition 4 obviously implies 3. Since step 1.2 gives 3 if and only if 1, condition 4 is equivalent to 1 as well.

Therefore 1, 2, 3, and 4 are equivalent.