When is this even possible (even for a dense graphs) $|E| = \Theta (|V|^2)$
You're absolutely right that $\Theta$ is the tightest asymptotic bound. But it's still asymptotic, and that means that we don't care about constant factors or terms of lower degree: when $n$ (or $v$ or whatever) gets big enough, the smaller terms become negligible.

In this case, $\frac{v(v-1)}{2} = \frac{1}{2} v^2 + \frac{-1}{2} v$. So removing the constant factors and lower terms leaves us with $v^2$.