Inconsistent definition of a vector in differential geometry? In the context of studying general relativity, I've seen two different definitions of a vector that seem inconsistent with one another.
The first says in some coordinate basis on a Riemannian manifold, a vector field is a derivation
$$V = V^\beta\frac{\partial}{\partial x^\beta}$$
using the Einstein summation convention. It is a mapping $V:C^\infty \rightarrow C^\infty$. For some smooth scalar field $\phi \in C^\infty$ we have
$$V(\phi) = V^\beta\frac{\partial \phi}{\partial x^\beta}$$
The second definition says a vector field is a ${1 \choose 0}$  tensor field. It operates on a one-form field $p$ to give a scalar field. In a coordinate basis
$$V(p) = V(p_\alpha \omega^\alpha) = p_\alpha V(\omega^\alpha) = p_\alpha V^\alpha \in C^\infty$$
where $\omega^\alpha$ are the basis one-forms and $p^\alpha$ are the components of the one-form field (they are scalar fields themselves).
In the second definition, as opposed to the first, we can pull the scalar field $p_\alpha$ out of the vector. How can these two definitions be consistent or even describe the same geometric object?
 A: You have $V$ induce a $p\mapsto V(p)$ from 1-form to a $C^{\infty}$ is because if you chose $p=p_{\alpha}\omega^{\alpha}$ and operate
\begin{eqnarray*}
V(p)&=&p(V)\\ 
&=&p_{\alpha}\omega^{\alpha}\left(V^{\mu}\frac{\partial}{\partial^{\mu}}
\right)\\
&=&p_{\alpha}V^{\mu}\omega^{\alpha}\left(\frac{\partial}{\partial x^{\mu}}
\right),\end{eqnarray*}
for each dual basis $\omega^{\sigma}$,
where, upon choosing these as the coordinated ones $w^{\alpha}=dx^{\alpha}$ then
\begin{eqnarray*}
V(p)&=&
p_{\alpha}V^{\mu}\omega^{\alpha}\left(\frac{\partial}{\partial x^{\mu}}\right)\\
\\
&=&p_{\alpha}V^{\mu}dx^{\alpha}\left(\frac{\partial}{\partial x^{\mu}}\right)\\
\\
&=&p_{\alpha}V^{\mu}\delta^{\alpha}_{\mu}\\
\\
&=&p_{\alpha}V^{\alpha}.
\end{eqnarray*}
A: The second definition is really a definition of a $1$-form and not a vector.
You first define a tangent vector (i.e., $(1,0)$-tensor) at a point $p$ to be a derivation $V_p: C^\infty \rightarrow \mathbb{R}$ such that if $f \in C^\infty$ has a critical point at $p$, then $V_p(f) = 0$. The set of all tangent vectors at $p$ is a vector space called the tangent space at $p$ and usually denoted $T_p$.
The dual vector space $T^*_p$ is called the cotangent space at $p$ and any element $\theta \in T_P^*$ is called a $1$-form, covector, or $(0,1)$-tensor. By definition, any element $\theta \in T^*_p$ is a linear function of $V \in T_p$. If $(\partial_1, \dots, \partial_n)$ is a basis of $T_p$ and $(\omega^1, \dots, \omega^n)$ is the corresponding dual basis, then given any $V = V^\alpha\partial_\alpha \in T_p$ and $1$-form $b = b_\alpha\omega^\alpha$, then by the definition of a dual basis,
$$
b(V) = = b_\alpha V^\alpha.
$$
