Let $U\subset\mathbb{R}^n$ be open and $f\colon U\to\mathbb{R}$ a continuously differentiable function. The the total differential $df$ is a differential $1$-form, i.e. $df(p)$ is a cotangent vector $df(p)\colon T_p(U)\to\mathbb{R}$, where $T_p(U)$ is the tangent space at $p\in U$.
Now, is there an analoug statement for the differential of a function $f\colon U\to\mathbb{R}^n$ with $U\subset\mathbb{R}^m$?
In this situation, the differential in $p$ is a map $df(p)\colon T_p(U)\to R^n$, isnt it? Is it also a differential $k$-form?
For your last question th answer is not, that $dp$ is not a $k$-form. $k$-forms are obtained with the exterior derivative that is different form the total derivative on your post.
The coincidence here between the exterior derivative and the total derivative is because $T\mathbb R$ is isomorphic to $\mathbb R$. But in general, you can compute the total derivative of a map between manifolds, while the exterior derivative only makes sense for $k$-forms. Namely, $k$-forms are antisymmetrizations of the tensor product of $1$-forms. You can get more information in Wiki for example, by searching differential form.
Suppose that $f:M\rightarrow N$ is a map between manifolds. Then, $df(p)$ maps to each tangent vector $X\in T_pM$ the vector field $Y\in T_{f(p)}N$ defined as the derivation $Y(g)=X(g\circ f)$.