This question is motivated by an exercise to prove that the Smale--Williams Solenoid is a hyperbolic set. The easiest way to do this is through invariant cones,but this requires computing the differential, and having never had a course in manifolds I fear my computation may be too simplistic.
We can think of the solid torus as the the product space $T=S^1\times D^2$, where $D^2$ is the closed disk. Consider the map $F:T\to T$ given by, for some $\lambda\in(0,1/2)$ $$F(\phi,x,y)=(2\phi,\lambda x+\frac{1}{2}\cos2\pi\phi,\lambda y+\frac{1}{2}\sin2\pi\phi).$$ As this map is contracting, I am only interested in the interior points. For any interior point $p$ I want to say that the differential $dF_p$ in these coordinates is simply the Jacobian $$dF_p= \begin{bmatrix} 2 & -\pi\sin2\pi\phi|_p& \pi\cos2\pi\phi|_p \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}, $$ but this seems far too simplistic. I don't think $\{\frac{\partial}{\partial\phi}|_p,\frac{\partial}{\partial x}|_p,\frac{\partial}{\partial y}|_p\}$ even serves as a basis for $\mathbb R^3$. Is this correct, or do I need to parameterize the torus with respect to the standard $\mathbb R^3$ coordinates (which I suspect the standard charts for $T$ are in) and then calculate the Jacobian of $F$ in these coordinates?
Simply write$$dF_p= \begin{bmatrix} 2 & -\pi\sin2\pi\phi& \pi\cos2\pi\phi \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}, $$ and use cones (exactly as you say, but without complicating).
The reason is that the solid torus is really a subset (meaning that it inherits its properties) of $\mathbb R^3$. And at the level that you describe it is really better not to complicate more (say in a Grassmannian or in something that really needs it).