I'm self-studying differential geometry and I wanted to convert the left hand side of $$\int_\gamma\nabla\phi\cdot d\mathbf r=\int_\gamma d\phi=\int_{\partial \gamma}\phi=\phi(\mathbf r_2)-\phi(\mathbf r_1)$$ to differential geometry language. In this expression we have a dot product between two vectors ($\nabla\phi$ and $d\mathbf r$), so I decided to write $d\mathbf r$ as a one-form acting on the vector $\nabla\phi$: \begin{align} \nabla\phi\cdot d\mathbf r=(dx\mathrm{dx}+dy\mathrm{dy}+dz\mathrm{dz})\left(\frac{\partial \phi}{\partial x}\frac{\partial }{\partial x}+\frac{\partial \phi}{\partial y}\frac{\partial }{\partial y}+\frac{\partial \phi}{\partial z}\frac{\partial }{\partial z}\right) \end{align} Here I used $\frac{\partial }{\partial x}$ as a basis vector in the x-direction and $\mathrm{dx}$ as basis one-form in the x-direction. Note that I used $dx$ for the "regular differential" and $\mathrm{dx}$ for the one-form. Since $\mathrm{dx_i}(\frac{\partial }{\partial x_j})=\delta^j_i$, this expression correctly(?) evaluates to $$\frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial z}dz$$ My confusion stems from the fact that I had to mix "regular differentials" and one-forms in the expression for $d\mathbf r$ and also that in the final expression I integrate over these regular differentials instead of one-forms. It feels illegal that I have the differentials and one-forms together. Any help?
To make it more clear: with "regular differentials" I mean scalar values $dx$ as you encounter in regular calculus, while with one forms I mean objects that eat vectors.
Suppose you have fixed a Cartesian coordinate system $(x, y, z)$ in your three-dimensional space and let $\mathbf{e}_x,\mathbf{e}_y,\mathbf{e}_z$ denote the corresponding unit vectors. The position vector $\mathbf{r}$ thus reads $$ \mathbf{r}=x\mathbf{e}_x+y\mathbf{e}_y+z\mathbf{e}_z, $$ and since $d\mathbf{e}_x=d\mathbf{e}_y=d\mathbf{e}_z=\mathbf{0}$, $$ d\mathbf{r}=dx\mathbf{e}_x+dy\mathbf{e}_y+dz\mathbf{e}_z.$$
On the other hand, the gradient reads $$ \nabla = \mathbf{e}_x\frac{\partial}{\partial x} +\mathbf{e}_y\frac{\partial}{\partial y}+\mathbf{e}_z\frac{\partial}{\partial z}, $$ so, using that $\mathbf{e}_x\cdot \mathbf{e}_x=1, \mathbf{e}_x\cdot\mathbf{e}_y=0$ (and so on), we conclude $$ \nabla \phi \cdot \mathbf{r} = \frac{\partial \phi}{\partial x}dx +\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial z}dz=d\phi,$$ as expected.
Remark 1. Note that I am avoiding writing things like $dx\mathrm{dx}$, which apart from being formally wrong, are also terrible choices of notation, for obvious reasons.
Remark 2. If you are a self-learner, you might want to check the book M.Itskov, "Tensor Algebra and Tensor Analysis for Engineers", Springer. https://link.springer.com/book/10.1007/978-3-319-16342-0 I liked it a lot as a student of mathematics.