Proof.
Part (1) is Lemma 31.5.9. Part (2) follows from (1) and Lemma 31.5.8.
Proof of part (3). To show the depth is $\geq 2$ it suffices to show that $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{S, s}}(\kappa (s), (f_*\mathcal{F})_ s) = 0$ and $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_{S, s}}(\kappa (s), (f_*\mathcal{F})_ s) = 0$, see Algebra, Lemma 10.72.5. Using the exact sequence $0 \to \mathfrak m_ s \to \mathcal{O}_{S, s} \to \kappa (s) \to 0$ it suffices to prove that the map
\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{S, s}}(\mathcal{O}_{S, s}, (f_*\mathcal{F})_ s) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{S, s}}(\mathfrak m_ s, (f_*\mathcal{F})_ s) \]
is an isomorphism. By flat base change (Cohomology of Schemes, Lemma 30.5.2) we may replace $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ and $X$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \times _ S X$. Denote $\mathfrak m \subset \mathcal{O}_ S$ the ideal sheaf of $s$. Then we see that
\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{S, s}}(\mathfrak m_ s, (f_*\mathcal{F})_ s) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ S}(\mathfrak m, f_*\mathcal{F}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(f^*\mathfrak m, \mathcal{F}) \]
the first equality because $S$ is local with closed point $s$ and the second equality by adjunction for $f^*, f_*$ on quasi-coherent modules. However, since $s \not\in f(X)$ we see that $f^*\mathfrak m = \mathcal{O}_ X$. Working backwards through the arguments we get the desired equality.
For the proof of (4), (5), and (6) we use flat base change (Cohomology of Schemes, Lemma 30.5.2) to reduce to the case where $S$ is the spectrum of $\mathcal{O}_{S, s}$. Then a nonzerodivisor $a \in \mathcal{O}_{S, s}$ deterimines a short exact sequence
\[ 0 \to \mathcal{O}_ S \xrightarrow {a} \mathcal{O}_ S \to \mathcal{O}_ S/a \mathcal{O}_ S \to 0 \]
Since $\mathcal{F}$ is flat over $S$, we obtain an exact sequence
\[ 0 \to \mathcal{F} \xrightarrow {a} \mathcal{F} \to \mathcal{F}/a\mathcal{F} \to 0 \]
Pushing forward we obtain an exact sequence
\[ 0 \to f_*\mathcal{F} \xrightarrow {a} f_*\mathcal{F} \to f_*(\mathcal{F}/a\mathcal{F}) \]
This proves (4) and it shows that $f_*\mathcal{F}/ af_*\mathcal{F} \subset f_*(\mathcal{F}/a\mathcal{F})$. If $b$ is a nonzerodivisor on $\mathcal{O}_{S, s}/a\mathcal{O}_{S, s}$, then the exact same argument shows $b : \mathcal{F}/a\mathcal{F} \to \mathcal{F}/a\mathcal{F}$ is injective. Pushing forward we conclude
\[ b : f_*(\mathcal{F}/a\mathcal{F}) \to f_*(\mathcal{F}/a\mathcal{F}) \]
is injective and hence also $b : f_*\mathcal{F}/ af_*\mathcal{F} \to f_*\mathcal{F}/ af_*\mathcal{F}$ is injective. This proves (5). Part (6) follows from (4) and (5) and the definitions.
$\square$
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