Show that "$\Gamma \models S \Rightarrow \Gamma \vdash S$" entails "if $\Gamma \nvdash P \And \sim P$ then $\Gamma$ is satisfiable" 
Show that "$\Gamma \models S \Rightarrow \Gamma \vdash S$" entails "if
  $\Gamma \nvdash P \And \sim P$ then $\Gamma$ is satisfiable"

I'm primarily confused with the notation being used here. In particular, I understand that "satisfiable" means there is at least one interpretation that makes $\Gamma$ true, but I'm not sure what $\vdash$ means nor what the statement $\Gamma \models S \Rightarrow \Gamma \vdash S$ means. The first part tells me that $\Gamma$ "makes true" $S$, or that there is no interpretation which makes $\Gamma$ true and $S$ false, but I don't understand how to interpret what comes after this (the "$\Rightarrow \Gamma \vdash S$" part). I'd really appreciate if someone could clear up this notation for me.

$\Gamma \models S$ means 'Every model of $\Gamma$ satisfies wff $S$ ' So given statement can be written as follows:

For all wff $S$ on the language of $\Gamma$, if every model of $\Gamma$ satisfies $S$ then $\Gamma$ proves $S$.

Formally, this statement is written as $\forall S[ (\forall \mathcal{M} :\mathcal{M}\models \Gamma \implies \mathcal{M}\models S)\implies \Gamma\vdash S]$. Take contraposition of this statement then we get

For all wff $S$ on the language of $\Gamma$, if $\Gamma$ does not prove $S$ then there is a model of $\Gamma$ that does not satisfy $S$.

(Formally it is written as $\forall S [\Gamma \nvdash S \implies (\exists \mathcal{M} :\mathcal{M}\models\Gamma \,\&\, \mathcal{M}\not\models S)]$.)
Note that if $\mathcal{M}\not\models S$ then $\mathcal{M}\models(\lnot S)$. Take $S=P\land \lnot P$ then you get desired proposition.
