I have two questions on chapter 4.1(pg. 34-35) of Tu's Textbook \textit{An Introduction to Manifolds}
Tu defines a differential 1-form on an open subset $U \subset \mathbb{R}^n$ to be a function $\omega$ that assigns a point $p$ a covector $\omega_p \in T^*_p(\mathbb{R}^n):$ \begin{align*} &\omega : U \rightarrow \bigcup_{p\in U} T^*_p(\mathbb{R}^n)\\ \end{align*} \begin{align*} p \mapsto \omega_p \in T^*_p(\mathbb{R}^n) \end{align*} He soon defines that a 1-form $df$, the differential of $f$ as: \begin{align*} (df)_p(X_p) = X_pf \end{align*} My confusion is, how is this definition of $df$ in accordance with $\omega$? Shouldn't $X_pf \in T_p(\mathbb{R}^n)$?, this is not the dual space $T^*_p(\mathbb{R}^n)$ as described in $\omega$.
Proposition 4.1, which is immediately after the definition, it states that {$dx^1\rvert_p,...dx^n\rvert_p$} is the basis for the cotangent space $T^*_p(\mathbb{R}^n)$, and the proof of it is rather short: \begin{align*} (dx^i)_p(\frac{\partial}{\partial x^j}\bigg\rvert_p)=\frac{\partial}{\partial x^j}\bigg\rvert_p x^i = \delta_j^i \end{align*} How does this close the proof? In order to show such is a basis, aren't we supposed to show that anything in the cotangent space is a linear combination of those, which are also independent?