I know that obviously the differential operator $D$ would be a differential form through the word differential. But in Spivak's Calculus on Manifolds he defines a $k$-form $w$ as $w(p) \in \Lambda^k (\mathbb{R}^n_p) $.
$ w(p) = \sum_{i_{1}<i_{2}<\dots<i_{k}} w_{i_{1}i_{2}\dots i_{k}}(p) \cdot[\phi_{i_{1}}(p)\wedge \phi_{i_{2}}(p)\wedge\dots\wedge \phi_{i_{k}}] $, where the $ \phi_{i} $ are the dual basis to the tangent space vectors. From this definition how does he get that for a scalar function $f$, $Df(p) \in \Lambda^1 (\mathbb{R}^n) $ ? Since $ \Lambda^k (\mathbb{R}^n_p) $ is a set of alternating tensors, does that mean that $Df(p)$ is also an alternating tensor?
A $1$-form is just a smooth choice of a linear functional on the tangent space at every point in the manifold. Note that every linear functional is alternating, but it doesn't really mean anything, since all permutations on a set of one vector are trivial.
Likewise, the derivative of a function is a linear functional on every tangent space. For $v\in T_pM$, $df_p(v)$ is a number, equal to the derivative of $f$ in the direction $v$.
In general, a $k$-form is indeed an alternating tensor that eats $k$ vector field and returns a function. A $1$-form thus eats one vector field and returns a function. Every tensor of type $(1,0)$ is alternating, obviously.