Proof of Majority is stablest in "reverse" in the MAXCUT hardness paper by Khot et al
"So, applying MIS on $g$"


To apply the Majority is Stablest theorem, you need to apply it to a non-negative parameter $\rho'\in[0,1)$ (read the statement of the theorem). Since in Proposition 7.3 the parameter $\rho\in(-1,0]$ is non-positive, this means you here apply it to $\rho' \stackrel{\rm def}{=} -\rho\in[0,1)$, giving $$\mathbb{S}_{-\rho}(g) \leq 1-\frac{2}{\pi}\arccos(-\rho)+\epsilon =  -1+\frac{2}{\pi}\arccos(\rho)+\epsilon$$ 
using that $\arccos x+\arccos(-x) = \pi$ for all $x$ (This may very well be the part you are missing. $\arccos$ is not an odd function).
Then
$$\mathbb{S}_{\rho}(f) \geq -\mathbb{S}_{-\rho}(g)  \geq -\left(-1+\frac{2}{\pi}\arccos(\rho)+\epsilon\right) = 1-\frac{2}{\pi}\arccos(\rho)-\epsilon$$ 
as stated.