Momentum a cotangent vector Apparently one identifies the configuration space in physics often with a manifold $M$. The tangent bundle $TM$ is then the space of all possible positions and velocities.
Furthermore, many sources seem to claim that $T^*M$ can be regarded as the phase space, where $(q,p) \in T^*M$ satisfies by definition that $p \in T_q^*M.$
Again by definition this means that $p:=\partial_2L$ takes velocities as arguments and is linear(!) in them. Unfortunately, I don't see from the definition of the momentum by the Lagrangian why this should be a linear functional. So something is confusing me here.
 A: The Lagrangian is a function on the tangent bundle $L:TM\rightarrow\mathbb{R}$. Given a point $q\in M$ and a Lagrangian, we can define a function $L_q:T_qM\rightarrow \mathbb{R}$ using the simple formula $L_q(v_q)=L(v_q)$, where $v_q\in T_qM$ is a tangent vector at $q\in M$. Notice that $L_q$ is a mapping between the vector spaces $T_qM$ and $\mathbb{R}$. We may therefore consider the Frechet derivative $DL_q:T_qM\rightarrow L(T_qM,\mathbb{R})$, where $L(T_qM,\mathbb{R})$ is the space of linear maps between $T_qM$ and $\mathbb{R}$. Notice that $DL_q$ is essentially the partial derivative of $L$ in the velocity direction. 
Here is how the cotangent bundle comes in. The set $L(T_qM,\mathbb{R})$ is precisely the cotangent space at $q$, i.e. $L(T_qM,\mathbb{R})=T^*_qM$. Therefore the Frechet derivative of $L_q$ is a map of the form $DL_q:T_qM\rightarrow T^*_qM$. This observation can be used to construct a mapping of $TM$ into $T^*M$ given by $v_q\mapsto (DL_q)(v_q)\equiv p_q\in T^*_qM$, which is known as the Legendre transform.
Does this help?
A: $p$ is the differential of L with respect to the second variable ($\dot q$), so it represents a linear functional on the tangent space at $q, \dot q$), given by
$$ (u,v)\to \frac{d}{dt} L(q,\dot q+tv)|_{t=0}.$$
