Basis for the cotangent space $T^\ast_p M$ and the maps $dx^i$ 
Let $M$ be a manifold and $p \in M$, then we have the tangent space $T_pM$ that has a basis $\left\{ \left(\frac{\partial}{\partial x^i}\right)_p \right \}$. The cotangent space $T^\ast_p M$ has then a basis $\{ (dx^i)_p \}$.

I'm trying to figure out why the cotangent space has the proposed base. If $V$ is a finite dimensional vector space with base $\{e_1, \dots e_n \}$, then $V^\ast$ has a basis $\{f^1, \dots f^n \}$, where $f^i : V \to \Bbb R$ and $f^i(e_j) = \delta_i^j$.
In the cotangent space case $\{ (dx^i)_p \}$ would be a basis if $$(dx^i)_p\left(\frac{\partial}{\partial x^j}\right)_p = \delta_i^j.$$
Now $(dx^i)_p : T^\ast_p M \to \Bbb R$ is just any linear map so how can I know anything about $(dx^i)_p\left(\frac{\partial}{\partial x^j}\right)_p$ without defining what $(dx^i)_p$ does first?
I'm told that $dx^i$ is "just a notation", but it should apparently be defined to be something in order for this to work out?
 A: Given local coordinates $(x^1,\ldots,x^n)$ on an open subset $U \subset M$, there are essentially two ways to define the coframe $(dx^1,\ldots,dx^n)$ on $U$:

*

*For $i\in \{1,\ldots,n\}$, we have a smooth function $x^i \colon U \to \Bbb R$.
Its differential $dx^i$ is a $1$-form on $U$.
With this definition, it is not a priori obvious that $(dx^1,\ldots,dx^n)$ is the dual coframe of $(\partial/\partial x^1,\ldots,\partial/\partial x^n)$.
But here is the reason: the integral curve of $\partial/\partial x^j$ through $p\in U$ is the curve given in coordinates by $t\mapsto (x^1(p), \ldots, x^j(p)+t,\ldots,x^n(p))$.
From this, you can easily derive that $dx^i(\partial/\partial x^j) = \delta^i_j$.

*The second way is the following: it is defined as the dual coframe of $(\partial/\partial x^1,\ldots,\partial/\partial x^n)$ pointwisely by the relation $dx^i_p(\partial/\partial x^j|_p) = \delta^i_j$ for $p\in U$.
Here, the $d$ in $dx^i$ is just a notation, and it is not a priori obvious that $dx^i$ is the differential of the smooth function $x^i$.
But the proof is not that complicated: it looks like that mentionned in the first point.

