I was reading an example from Lee's book to try understanding the two body problem (see below). Since the forces are conservative, there exists a $U : Q \to \mathbb{R}$ such that $F=dU$, so I understand that $F$ is a one-form, but when we have $F_q(v_q)$, $v_q \in T_qQ$ is a velocity, no? What does it mean to evaluate the force into a velocity. There is something I am missing here...
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Allthough i see no mentionings of $F(v)$ in the text cited by you, the physical meaning of $F(v)$ is the power consumed by the system, when forced into a movement according to $q$ with $\dot{q} = v$. This is easily derived from the formula for the work $$ W = \int\limits_s F_q(dq) = \int\limits_0^T F_{q(t)}(\dot{q}) dt $$ performed on the system driven to a motion along a curve $s$: $\partial_TW = F_{q(t)}(\dot{q})$.
One-forms eat vectors and spit out scalars. That's what they do.
In the special case of the one-forms $dU$ for scalar functions $U$, given a vector $v_q$, the definition of $d$ gives us that $dU_q(v_q)$ is equal to $v_q(U)$. Which is to say, the directional derivative of $U$ at $q$ along $v_q$. This is true at any specific point, and it's true for vector / scalar fields.
The most apparent physical interpretation of this is that we might want to find the work done by the force on an object following a path $\gamma$, parameterised as $\gamma:[a,b]\to X$ (where $X$ is your manifold). This work is given by the integral $$ \int_\gamma F=\int_\gamma dU\\ =\int_a^b v_{\gamma(t)}(U)dt $$ where $v_{\gamma(t)}$ is the velocity / tangent vector of $\gamma$ at time $t$.