To expand on my comment, consider $T:L^2\to L^2$ bounded self-adjoint operator, a necessary condition is that $\not\exists f\in \ker T$ and $g\in (\ker T)^\perp$ such that $f Tg \not\equiv 0$. Otherwise $\|Tf g\| = 0$ is not comparable to $\|f Tg\| \neq 0$.
This is, of course, not sufficient, since if you let $\widehat{Tf} = e^{-\lfloor\xi\rfloor}\hat{f}(\xi)$. This operator is bounded on $L^2$, and has no kernel. But if you let $\hat{f}$ be the ball of radius 1 and $\hat{g}_n$ be the annulus of inner radius $n$ and outer radius $n+1$, you will still have no uniform constant comparing $\|fTg\|$ with $\|Tfg\|$.
A sufficient condition, OTOH, at least for functions in $L^2$, you can get via the spectral theorem. If $T$ is a bounded linear operator, consider the measure space $(X,\Sigma,\mu)$ and the isomorphism $\Phi$ from $L^2$ to $L^2_\mu(X)$ such that $T$ is represented by $f\mathrm{d}\mu$. If $\Phi$ is such that whenever $u,v\in L^2_\mu(X)$ have disjoint essential support then so do $\Phi^{-1}u,\Phi^{-1}v$, we can conclude that (if I am not mistaken) that $\|Tf g\| = \|f Tg\|$. But this is a very artificial and strong condition. And I don't know if this is at all easy to verify in applications.