The (velocity) phase space of a double pendulum can be seen as the tangent bundle of its configuration space ($\mathcal{S}^1\times\mathcal{S}^1$), that is: $\mathcal{S}^1\times\mathcal{S}^1\times\mathbb{R}^2$ [proof : $T(M\times N)=T(M)\times T(N)$, if I'm correct], which is homeomorphic to $\mathcal{T}^2\times\mathbb{R}^2$.
In the case of a simple pendulum, the tangent bundle $\mathcal{S}^1\times\mathbb{R}$ is homeomorphic to a cylinder, of course easy to visualize as a surface in $\mathbb{R}^3$.
Is there somehow an clever way to visualize the tangent bundle of the torus $\mathcal{T}^2$?
Sure. The tangent bundle to $S^1 \times S^1$ is simply the product of the tangent bundles of the two factors. Since each of these is a cylinder, a point of the object you're interested in can be thought of as a pair of points, one on each of two cylinders. I'm not certain that this helps, but...
...to be honest, if I wanted to visualize the tangent bundle to the torus, I might say "Hey, it's naturally identified with the phase space of a double-pendulum!" and consider that I had a good answer. :)
There's an interpretation for the two cylinder points: each represents the angle for one arm of the pendulum, with the "long" coordinate (along the cylinder axis) indicating angular velocity. You can skew this, and make one point represent the first arm, and the other represent the difference of the first arm from the second, in which case the "long" coordinate represents the difference of angular velocities. And by choosing one of many other diffeomorphisms of the torus, you can build lots of others....none seem particularly helpful to me, though.