Understanding of Differential one Form [Spivak] I am using Spivak Calculus on Manifolds to understand differential form. I understand that a k-form is defined to be the function sending a point to this point paired with an alternating k-tensor. I have confusion of the following discussion:

If $f :\mathbb R^n \to \mathbb R$ is differentiable, then $Df(p) \in \Lambda^1 (\mathbb R^n)$. By a minor modification we therefore obtain a 1-form $df$, defined by
  $$ df(p)(v_p) = Df(p) (v).$$

I wonder what does $Df(p)$ means here. Why does the derivative of $f$ at a point $p$ become a 1-tensor instead of a point in $\mathbb{R^n}$? Does the RHS mean $Df(p)\cdot (v)$?

Perhaps a little late to the party, but if you read through Chapter $2$ again, Spivak defined the derivative of a function at a point to be a linear transformation. So, if $f$ is a function from $\mathbb{R^n} \to \mathbb{R}$, and if $f$ is differentiable at a point $p \in \mathbb{R^n}$, $Df(p)$ was defined to be a linear transformation $Df(p): \mathbb{R^n} \to \mathbb{R}$. So, $Df(p) \in \mathcal{L}(\mathbb{R^n}, \mathbb{R}) = \Lambda^1(\mathbb{R^n})$. Hence it is a $1$-tensor on $\mathbb{R^n}$, simply by definition.
So in the equation
\begin{equation}
df(p)(v_p) = Df(p)(v),
\end{equation}
perhaps it might be useful to add another pair of brackts too see how things are being evaluated:
\begin{equation}
\left(df(p) \right)(v_p) = \left( Df(p) \right)(v).
\end{equation}
On the RHS, you are evaluating the  linear transformation $Df(p): \mathbb{R^n} \to \mathbb{R}$ at the point $v \in \mathbb{R^n}$.
