Reducing MAX-2SAT to Vertex Cover? In A simplified NP-complete MAXSAT problem, a reduction is given from Min Vertex Cover to MAX-2SAT by replacing each each vertex $x_i$ by a single-variable clause, and each edge by a two-variable clause:

\begin{align}
 \Phi = \left(\bigwedge_{i=1}^n x_i\right) \wedge \left(\bigwedge_{\lbrace i,j\rbrace \in E} (\overline{x}_i \vee \overline{x}_j)\right) 
\end{align}

This basically makes sense to me, because the QUBO version of Vertex Cover is to maximize:
\begin{align}
L = \sum_{i=1}^N x_i -  \sum_{\lbrace i,j\rbrace \in E} x_ix_i
\end{align}
and QUBO can be converted to MAX-2SAT quite simply.

However, I would to know how the reverse transformation works. 

How do you go from MAX-2SAT to Vertex Cover?

I don't actually know if this is an unsolved problem or not, but I figure it shouldn't be since they are both NP-Complete. Would it be as simple/tedious as trying to force an arbitrary MAX-2SAT instance into the same form as $\Phi$? I don't know if that can be done though.
You can reduce MAX-2SAT to Satisfiability (using Cook-Levin), which can then be reduced to Vertex Cover (Folklore).