Sum-of-Squares Certificates
I see the confusion, but I think the document you provided pretty well explains what is meant: solving MAXCUT on a graph $G$ is equivalent to finding the smallest value of $c$ such that $c-f_G(x)\geq 0$ for every $x\in \{-1,1\}^n$. As you write, it is trivially true that $c^*=\mathsf{OPT}(G)$ is the optimal value where this holds by definition, but for one, you want to determine the value of $\mathsf{OPT}(G)$ explicitly, and equally as important, there will not in general be a degree 2 sum-of-squares proof for any $c$ better than $\mathsf{OPT}(G)/.878$ (this doesn't rely on UGC or anything fancy like that; if I remember right, there are known hard examples that basically embed vectors in a high-dimensional sphere with edges that emulate where the GW algorithm has a tough time in the rounding and uses the isoperimetric inequality on the sphere to argue about the real optimal value). The point of doing the degree 2 sum-of-squares algorithm is that there for sure exists a degree 2 sum-of-squares certificate of this polynomial inequality over $\{-1,1\}^n$ for any $c\geq\mathsf{OPT}(G)/.878$ (which is what the work in the literature you reference establishes), so by doing binary search, you can algorithmically determine this looser quantity efficiently.