When writing $df$, for a differentiable function $f:\mathbb{R}^{n} \to \mathbb{R}$, I know rigorously speaking, it is a one form, such that $df(p)(v_{p}) = Df(p)(v)$, where $Df(p)$ is the linear transformation for the derivative, and $Df(p) \in \Lambda^{1}(\mathbb{R}^{n})$ ( $\Lambda^{i}(V)$ , means the alternating i tensors on V. $dx^{i}$ then denotes $d\pi^{i}$, where $\pi^{i}$ is the projection map to the ith entry.)
Now in matrix calculus, people write dg, where $g:\mathbb{R}^{n} \to \mathbb{R}^{n}$, is this essentially just the column vector such that ith entry is $dg^{i}$? Now, for $k:\mathbb{R}^{n \times n} \to \mathbb{R}^{n\times n}$, is dk just the matrix where i,j th entry is $dk^{i,j}, \text{the projection to the i,j th entry}$, where $k^{i,j}: \mathbb{R}^{n \times n} \to \mathbb{R}$? Is my interpretation correct?
My understanding of differential forms is quite elementary, basically from reading Calculus on Manifolds, if this interpretation is correct, I also wonder does there exist a coordinate-free version, or in general, for a function $f:V \to W$, does $df$ make sense?
Thank you for helping!
Edit: The comments seem to suggest that I can think about df as just Df, but I still wonder, is my interpretation equivalent to the definition of vector-valued 1 forms in this case? Though no coordinates are nice theoretically, thinking $df, f:\mathbb{R}^{n \times n} \to \mathbb{R}^{n \times n}$ as a matrix is easier to work with.
In fact, in one source I'm reading the author defined $df (f: \mathbb{R}^{n\times n} \to \mathbb{R})$ as $\sum_{i=1}^{m}\sum_{j=1}^n \frac{ \partial f }{ \partial X_{ij} }dX_{ij} = tr\left( \frac{ \partial f }{ \partial X }^{\top} dX\right)$, I don't how $dX_{ij}$ would make sense then