I'm having some trouble seeing
$G=\langle S|R\rangle$ is finitely presented and simple, then if the world in free group $F(S)$, $w\neq e$ in $G$, $\langle\langle w\cup R \rangle\rangle = F(S)$
Symbol $\langle \langle \cdot \rangle \rangle$ denotes normal closure of the set of relations.
Since $G$ is simple, $w\neq e$, we would have $\langle \langle w \rangle \rangle = G$ since only normal set containing $w$ is $G$ itself. But why does this say that $\langle\langle w\cup R \rangle\rangle = F(S)$?
$G$ is only $F(S)/\langle\langle R \rangle\rangle$ so I don't see any easy connection here.
Since $w\neq e$ in $G$, therefore, $w\not\in\langle\langle R\rangle\rangle$ in $F(S)$. Thus, $\langle\langle w\cup R\rangle\rangle$ is a normal subgroup of $F(S)$ properly containing $\langle\langle R\rangle\rangle$. It follows that its image under the canonical homomorphism $F(S)\to G$ is a non-trivial normal subgroup of $G$. Since $G$ is simple, it must be all of $G$, so $\langle\langle w\cup R\rangle\rangle = F(S)$. This follows from the Correspondence Theorem, which establishes a one-to-one correspondence between the normal subgroups of $G$ and the normal subgroups of $F(S)$ containing $\langle\langle R\rangle\rangle$.