I'm not sure why you're talking about codimension $1$ components of $S$ since the smooth locus $S$ is actually dense in $\mathcal{M}_g$.

Anyway, the fundamental group of the locus $S$ of smooth points is the mapping class group $Mod_g$ of the surface $\Sigma_g$. To see this, recall that $\mathcal{M}_g$ is the quotient of Teichmuller space $\mathcal{T}_g$ by the action of $Mod_g$. Let $\widetilde{S}$ be the preimage of the smooth locus in $\mathcal{M}_g$ under the projection $\mathcal{T}_g \rightarrow \mathcal{M}_g$. The action of $Mod_g$ on $\mathcal{T}_g$ preserves $\widetilde{S}$, and moreover the restriction of the action of $Mod_g$ to $\widetilde{S}$ is free (the only fixed points of the action of $Mod_g$ on $\mathcal{T}_g$ come from curves with automorphisms). It follows that $\widetilde{S} \rightarrow S$ is a regular cover with deck group $Mod_g$. To prove that the fundamental group of $S$ is $Mod_g$, it is enough to prove that $\widetilde{S}$ is $1$-connected. For this, recall that the locus of curves with automorphisms has complex codimension $2$ in $\mathcal{M}_g$ as long as $g \geq 3$. This implies that the complement in $\mathcal{T}_g$ of $\widetilde{S}$ also has complex codimension $2$ and thus real codimension $4$, which implies in particular that the inclusion $\widetilde{S} \hookrightarrow \mathcal{T}_g$ induces an isomorphism on fundamental groups. The desired result now follows from the fact that $\mathcal{T}_g$ is contractible.

I'm not sure why you're talking about codimension $1$ components of $S$ since the smooth locus $S$ is actually dense in $\mathcal{M}_g$.

Anyway, the fundamental group of the locus $S$ of smooth points is the mapping class group $Mod_g$ of the surface $\Sigma_g$. To see this, recall that $\mathcal{M}_g$ is the quotient of Teichmuller space $\mathcal{T}_g$ by the action of $Mod_g$. Let $\widetilde{S}$ be the preimage of the smooth locus in $\mathcal{M}_g$ under the projection $\mathcal{T}_g \rightarrow \mathcal{M}_g$. The action of $Mod_g$ on $\mathcal{T}_g$ preserves $\widetilde{S}$, and moreover the restriction of the action of $Mod_g$ to $\widetilde{S}$ is free (the only fixed points of the action of $Mod_g$ on $\mathcal{T}_g$ come from curves with automorphisms). It follows that $\widetilde{S} \rightarrow S$ is a regular cover with deck group $Mod_g$. To prove that the fundamental group of $S$ is $Mod_g$, it is enough to prove that $\widetilde{S}$ is $1$-connected. For this, recall that the locus of curves with automorphisms has complex codimension $2$ in $\mathcal{M}_g$ as long as $g > 3$. This implies that the complement in $\mathcal{T}_g$ of $\widetilde{S}$ also has complex codimension $2$ and thus real codimension $4$, which implies in particular that the inclusion $\widetilde{S} \hookrightarrow \mathcal{T}_g$ induces an isomorphism on fundamental groups. The desired result now follows from the fact that $\mathcal{T}_g$ is contractible.