Why are $(q,p)$ in cotangent bundle instead of tangent bundle? In the book of Frankel, The geometry of Physics, at page 55, he states that

[...] Thus, 
$$p_i^{U} = \sum_j p_j^V (\frac{ \partial q_i^{U}}{\partial q_i^{V} }
 ),$$
and so the $p’s$ represent then not the components of a vector on the
  configuration space $M^n$ but rather a covector. The $q’s$ and $p’s$
  then are to be thought of not as local coordinates in the tangent
  bundle but as coordinates for the cotangent bundle.

However, I cannot understand why $p's$ represent a covector. I mean, I don't see any connection between the result about how does p's transform for a given coordinate patch to another, with why p's are convectors.
Edit:
In this question/comment, a similar thing is asked/answered, but none of the answers gives a clear & complete answer. For example, I don't understand why $p$ is a covector ?
 A: In order to understand the situation clearly, let us analyse it in more neutral environment, where the terminology and the physical properties of the variables does not block our view.
Let $M$ be a smooth manifold, and $f: M \to \mathbb{R}$ be smooth map. Later, $M$ be our tangent bundle in question, and $f$ be a Lagrangian. Let $y \in M$ be any point, and $\phi: y\in U\subseteq M \to \mathbb{R}^n$ be local chart around $y$ s.t $\phi(y) = (x_1(y), ..., x_n(y)) = (0,...,0)$. 
Let use define 
$$g_i (y) = \frac{\partial (f)}{ \partial x_i}|_\vec 0 := \frac{d }{dt }((f \circ x_i^{-1})(t))|_{t=0},  $$
where $x_i^{-1}: J \subseteq \mathbb{R} \to U$ s.t $\phi \circ x_i^{-1} = id_J$.
($g_i$ later be the momenta $p_i$)
However, since $\mathbb{R}$ is itself a smooth manifold, 
$$=df(p, \frac{\partial}{\partial x_i } ),$$
Thus, $g_i (\_) = df(\_, \frac{\partial}{\partial x_i } ) : M \to \mathbb{R}$.
Now, the above analysis hold for any smooth manifold $M$. In particular, for any smooth manifold $N$, let $M:= TN$, and $f: = L$, $g_i:= p_i$. Hence,
$$p_i: TN \to \mathbb{R},$$
thus, $p_i \in T^* N.$
