I'm reading Loring Tu's "An Introduction to Manifolds", and the definition given for the 1-form $df$ of a $C^\infty$ function $f$ isn't clicking for me. He writes on page 34,
A covector field or a differential 1-form on an open subset $U$ of $\mathbb{R}^n$ is a function $\omega$ that assigns to each point $p$ in $U$ a covector $\omega_p\in T^\ast_p(\mathbb{R}^n)$, $$ \omega: U\rightarrow \bigcup_{p\in U} T^\ast_p(\mathbb{R}^n)$$ $$ p\mapsto \omega_p\in T^\ast _p(\mathbb{R}^n).$$ Note that in the union $\bigcup_{p\in U} T^\ast_p(\mathbb{R}^n)$, the sets $T^\ast_p(\mathbb{R}^n)$ are all disjoint. We call a differential 1-form a 1-form for short. From any $C^\infty$ function $f:U\rightarrow \mathbb{R}$, we can construct a 1-form $df$, called the differential of $f$, as follows. For $p\in U$ and $X_p\in T_p U$, define $$ (df)_p(X_p)=X_p f.$$
Note that $T_p(\mathbb{R}^n)$ denotes the tangent space at a point $p\in U$ (i.e., all the vectors emanating from $p$), and $T^\ast_p(\mathbb{R}^n)$ denotes the dual of the tangent space $T_p(\mathbb{R}^n)$. Also, $X_p$ denotes a given vector in $T_p(\mathbb{R}^n)$. Now, after proving that $T_p(\mathbb{R}^n)$ is isomorphic to the set of all derivations at $p$, and hence has basis $\left.\partial/\partial x^1\right|_p,\dots,\left.\partial/\partial x^n\right|_p$ he remarks that all $X_p\in T_p(\mathbb{R}^n)$ can be written as $$ X_p=\sum_i a^i(p) \left.\frac{\partial}{\partial x^i}\right|_p $$ for $a^i(p)\in \mathbb{R}$. Then, he gives the following definition on page 16: $$ X_pf=\sum a^i(p) \left.\frac{\partial f}{\partial x^i}\right|_p. $$
Now that we've got all the notation and definitions out of the way, onto my main question: He says that a 1-form is a function which maps a point $p$ to a linear map $\omega_p:T_p(\mathbb{R}^n)\rightarrow \mathbb{R}$. Okay, fine. But in the definition of the differential of $f$, it is not clear to me which linear map in particular $df$ sends a point $p$. In other words, what is the precise image of a point $p\in U$ under the map $df$ - In the classic function notation "$g(x)$", what exactly is $df(p)$? In his definition it would appear that $df$ sends the point $p$ to $X_pf$. But $X_p$ is just any vector in $T_p(\mathbb{R}^n)$. How is $X_p$ chosen exactly? Is a particular $X_p$ given whenever $df$ is mentioned, so that $X_p f$ is unambiguous?
Moreover, isn't the image of a point $p$ under $df$ suppose to be a linear function from $T_p(\mathbb{R}^n)$ to $\mathbb{R}$, by definition of a 1-form? Because the map $X_pf$ is a map from $U$ to $\mathbb{R}$. So the image of $p$ under $df$ isn't even an element of $T^\ast_p(\mathbb{R}^n)$ based on this definition of $df$, but instead an element of $C^\infty(U)$. I have a feeling that I am just largely misunderstanding the notation here.
Please keep in mind: I have done absolutely nothing with regards to manifolds thus far, and am only in the first chapter of this textbook.