I am studying the theory of smooth manifolds. Currently, I am quite stuck in an example about the one-form defined by the differential of a smooth map. I'll write first the notation I'm using.
Let $M$ be a smooth manifold with atlas $\{(U,\varphi)\}$, let $T^*M=\bigcup_{p\in M}T^*_pM$ where $T^*_pM=\{f:T_pM\to\mathbb{R} \text{ linear}\}$. One can give $T^*M$ structure of smooth manifold via the charts $(T^*U,\tilde{\varphi}^*)$ where $T^*U=\bigcup_{p\in U}T^*_pM$ and $\tilde{\varphi}^*:T^*U \to \varphi(U)\times(\mathbb{R}^m)^*$ is given by $\tilde{\varphi}^*(\omega)=(\varphi(p),((\tau_\varphi^p)^{-1})^*(\omega))$ with $\tau_\varphi^p$ the isomorphism from $T_pM$ to $\mathbb{R}^m$.
Now, we define a one-form to be a differentiable section of the canonical map $\pi^*:T^*M\to M$, i.e. a map $\omega:M\to T^*M$ such that every point in $M$ defines an element $\omega_p \in T_p^*M$.
Given a chart $(U,\varphi)$ of $M$ and a one-form $\omega:M\to T^*M$, we can look at how does $\omega$ work in $U$.
$$\require{AMScd} \begin{CD} U @>{\omega}>> T^*U\\ @V{\varphi}VV @V{}VV \\ \varphi(U)@>{\tilde{\varphi}^*\circ\omega\circ\varphi^{-1}}>> \varphi(U)\times(\mathbb{R}^m)^*; \end{CD}$$
Now, $\tilde{\varphi}^*\circ\omega\circ\varphi^{-1}(x)=(x,g^U(x))$ and $g^U:\varphi(U)\to(\mathbb{R}^m)^*$ is called the local component of $\omega$ in $U$.
We can finally talk about the differential. Let $f:M\to\mathbb{R}$ be a smooth map. We define the one-form $$p\in M \mapsto (df)_p : T_pM \to T_{f(p)}\mathbb{R}\cong \mathbb{R}$$ My notes say that its local components are:
$$(\nabla f)^U_{\varphi(p)}=\left(\dfrac{\partial(f\circ\varphi^{-1})}{\partial x_1}\bigg\vert_{\varphi(p)},\dots,\dfrac{\partial(f\circ\varphi^{-1})}{\partial x_m}\bigg\vert_{\varphi(p)}\right)$$
So this might be a linear map from $\mathbb{R}^m$ to $\mathbb{R}$. But... this is a $\mathbb{R}^m$ vector... how is this the component of the differential one-form?
Then it says that if $e_i^p$ are the canonical basis of $T_pM$, and $(e_i^p)^*$ of $T_p^*M$, the differential $df$ defines a one-form in $(U,\varphi)$:
$$(df)_p=\dfrac{\partial(f\circ\varphi^{-1})}{\partial x_1}\bigg\vert_{\varphi(p)}(e_1^p)^*+\dots+\dfrac{\partial(f\circ\varphi^{-1})}{\partial x_m}\bigg\vert_{\varphi(p)}(e_m^p)^*$$
Now I could understand that this is the local expression of the one-form. But then, what is $(\nabla f)^U_{\varphi(p)}$? THe things that confuses me the most is that when checking the compatibility relation of local components, $g^V(\psi(p))(y)=g^U(\varphi(p))(J_{\psi(p)}(\varphi\circ\psi^{-1}))^t$, the notes use $(\nabla f)^U_{\varphi(p)}$ instead of the expression in $(e_i^p)^*$.
I'd like to understand this example because it seems quite important. The differential of coordinate maps $\varphi_i$ will correspond to vectors $(e_i^p)^*$ and this will be used to write one-forms comfortably.
Sorry for the extended question, I wanted to write it to explain my notations, and if you see any sign of me not understanding the topic properly, please tell me.
Note that $df_p : T_pM \to T_p\mathbb{R}\cong\mathbb{R}$ is linear, so we can view it as an element of $T^*_pM$. The basis $\{(e_i^p)^*\}$ induces an isomorphism $T_p^*M \to \mathbb{R}^m$ given by
$$a_1(e_1^p)^* + \dots + a_m(e_m^p)^* \mapsto (a_1, \dots, a_m).$$
Under this map, $df_p \mapsto (\nabla f)^U_{\varphi(p)}$; that is, $(\nabla f)^U_{\varphi(p)}$ are the components of $df_p$ with respect to the basis $\{(e_i^p)^*\}$.