I'm familiar with the ''coordinate view'' of Lagrangian and Hamiltonian mechanics where if $\pmb{q}=(q^1,\dots, q^n)\in\mathbb{R}^n$ are any n generalized coordinates and $L(\pmb{q},\dot{\pmb{q}})$ is the Lagrangian for whatever system we care about, then we define the conjugate momenta (coordinates) as $\pmb{p} = (p_1,\dots, p_n)= \frac{\partial L}{\partial \dot{\pmb{q}}}$ and the Hamiltonian as $H(\pmb{q},\pmb{p}) = p_i \dot{q}^i - L$ (with the RHS expressed in terms of $(\pmb{q},\pmb{p})$).

But I am recently learning the geometric view of things where $(\pmb{q},\dot{\pmb{q}})$ is just one of infinite possible coordinate representations of $(\text{x},\mathbf{v})\in TQ$, where $\text{x}\in Q$ and $\mathbf{v}_{\text{x}}= \frac{\text{d}}{\text{d}t}\text{x}\in T_{\text{x}}Q$, where $Q$ is some n-dimensional configuration manifold, and $L\in \mathcal{F}(TQ)$ a smooth function on the tangent bundle. The momenta is then a 1-form defined as some fiber derivative or Frechet derivative (I don't know what those words mean) as:

$$ \mathbf{p} := \mathbf{d}_2 L(\text{x},\mathbf{v}) \; \in \Omega^1(Q) \qquad\qquad (1) $$

where $\mathbf{d}_2$ is differential with respect to the second argument. The coordinate representation of the above is simply the familiar $$ p_i = \frac{\partial L}{\partial \dot{q}^i} \qquad\qquad (2) $$

Question: So equation (1) defines the actual momenta, a 1-form $\mathbf{p}\in\Omega^1(Q)$, and equation (2) just gives the components/coordinates, $\pmb{p}\in\mathbb{R}^n$, of this 1-form in some basis (right?). But what is this basis, exactly? If we write $\mathbf{p} = \frac{\partial L}{\partial \dot{q}^i} \pmb{\epsilon}^i$ what are the basis 1-forms, $\pmb{\epsilon}^i$? I know the coordinates $q^i$, regarded as functions, give coordinate basis vectors $\pmb{\partial}_i\in\mathfrak{X}(Q)$ and basis 1-forms $\mathbf{d}q^i \in \Omega^1(Q)$. The velocity is then $\mathbf{v}=\dot{q}^i\pmb{\partial}_i$ and I feel like the momenta should be $\mathbf{p}=p_i\mathbf{d}q^i = \frac{\partial L}{\partial \dot{q}^i}\mathbf{d}q^i $. But from equations (1) and (2) it looks like:

$$ \mathbf{p} := \mathbf{d}_2 L(\text{x},\mathbf{v}) \; \overset{?}{=} \; \frac{\partial L}{\partial \dot{q}^i} \mathbf{d} \dot{q}^i $$

Is the second equality above correct? If not, why? Perhaps my interpretation of $\mathbf{d}_2 L$ is wrong? If the above is correct, what is the relation between $\mathbf{d}\dot{q}^i$ and $\mathbf{d}q^i$? It would seem $\mathbf{d}\dot{q}^i = \frac{\text{d}}{\text{d}t} \mathbf{d}q^i$ but it doesn't seem that this would form a basis for each $T_{\text{x}}^*Q$, in general.

I'm familiar with the ''coordinate view'' of Lagrangian and Hamiltonian mechanics where if $\pmb{q}=(q^1,\dots, q^n)\in\mathbb{R}^n$ are any $n$ generalized coordinates and $L(\pmb{q},\dot{\pmb{q}})$ is the Lagrangian for whatever system we care about, then we define the conjugate momenta (coordinates) as $\pmb{p} = (p_1,\dots, p_n)= \frac{\partial L}{\partial \dot{\pmb{q}}}$ and the Hamiltonian as $H(\pmb{q},\pmb{p}) = p_i \dot{q}^i - L$ (with the RHS expressed in terms of $(\pmb{q},\pmb{p})$).

But I am recently learning the geometric view of things where $(\pmb{q},\dot{\pmb{q}})$ is just one of infinite possible coordinate representations of $(\text{x},\mathbf{v})\in TQ$, where $\text{x}\in Q$ and $\mathbf{v}_{\text{x}}= \frac{\text{d}}{\text{d}t}\text{x}\in T_{\text{x}}Q$, where $Q$ is some n-dimensional configuration manifold, and $L\in \mathcal{F}(TQ)$ a smooth function on the tangent bundle. The momenta is then a 1-form defined as some fiber derivative or Frechet derivative (I don't know what those words mean) as:

$$ \mathbf{p} := \mathbf{d}_2 L(\text{x},\mathbf{v}) \; \in \Omega^1(Q) \tag{1} $$

where $\mathbf{d}_2$ is differential with respect to the second argument. The coordinate representation of the above is simply the familiar $$ p_i = \frac{\partial L}{\partial \dot{q}^i}. \tag{2} $$

Question: So equation (1) defines the actual momenta, a 1-form $\mathbf{p}\in\Omega^1(Q)$, and equation (2) just gives the components/coordinates, $\pmb{p}\in\mathbb{R}^n$, of this 1-form in some basis (right?). But what is this basis, exactly? If we write $\mathbf{p} = \frac{\partial L}{\partial \dot{q}^i} \pmb{\epsilon}^i$ what are the basis 1-forms, $\pmb{\epsilon}^i$? I know the coordinates $q^i$, regarded as functions, give coordinate basis vectors $\pmb{\partial}_i\in\mathfrak{X}(Q)$ and basis 1-forms $\mathbf{d}q^i \in \Omega^1(Q)$. The velocity is then $\mathbf{v}=\dot{q}^i\pmb{\partial}_i$ and I feel like the momenta should be $\mathbf{p}=p_i\mathbf{d}q^i = \frac{\partial L}{\partial \dot{q}^i}\mathbf{d}q^i $. But from equations (1) and (2) it looks like:

$$ \mathbf{p} := \mathbf{d}_2 L(\text{x},\mathbf{v}) \; \overset{?}{=} \; \frac{\partial L}{\partial \dot{q}^i} \mathbf{d} \dot{q}^i.\tag{3} $$

Is the second equality above correct? If not, why? Perhaps my interpretation of $\mathbf{d}_2 L$ is wrong? If the above is correct, what is the relation between $\mathbf{d}\dot{q}^i$ and $\mathbf{d}q^i$? It would seem $\mathbf{d}\dot{q}^i = \frac{\text{d}}{\text{d}t} \mathbf{d}q^i$ but it doesn't seem that this would form a basis for each $T_{\text{x}}^*Q$, in general.