Is the differential $\mathrm{d}\vec{r}$ a sensible mathematical object? When doing differential geometry, physicists often use
$$\mathrm{d}\vec{r} = \mathrm{d}x^i\space\vec{e}_i$$
for many different things. For instance, they define the holonomic basis $\{\vec{e}^{\space\prime}_a\}$ relative to a coordinate system $\{x'^a\}$ by imposing
$$\mathrm{d}\vec{r} = \mathrm{d}x'^{a}\space\vec{e}^{\space\prime}_a \implies \vec{e}^{\space\prime}_a=\frac{\partial\vec{r}}{\partial x'^a}$$
and they compute the quadratic form of the metric $\mathrm{d}s^2$ as $\mathrm{d}\vec{r}\cdot\mathrm{d}\vec{r}$.
Computing the differential of a vector field ($\vec{r}=x^i\vec{e}_i$, in this case) feels strange, as in differential geometry differentials are usually considered to be alternating $k$-forms, so it would only make sense to talk about the differential of a scalar field (aka its exterior derivative).
Not only that, the "true" definitions of holonomic bases and $\mathrm{d}s^2$ don't use this $\mathrm{d}\vec{r}$ at all.

EDIT: in fact, taking the derivative of $\vec{r}$, or any other vector field, is something we are not allowed to do in a general differentiable manifold without a connection, so we obviously wouldn't define a holonomic basis like that. A holonomic basis would basically be the basis formed by the tangent vectors $\partial/\partial x'^a$.

After thinking about it, I thought the differential of a vector field might just be
$$\mathrm{d}\vec{\varphi} = (\nabla_i\varphi^j)\space\vec{e}_j\otimes\mathrm{d}x^i,$$
so maybe $\mathrm{d}\vec{r} = \mathrm{d}x^i\space\vec{e}_i$ means $\mathrm{d}\vec{r} = \mathrm{d}x^i\otimes\vec{e}_i$? How is $\mathrm{d}\vec{r}$ rigorously defined, otherwise?
 A: You're correct. it's a tensor product. Abstractly, on a vector space $V$, $d\vec{r} \in V\otimes V^*$. Any element of this tensor space defines a map $V \rightarrow V$. For example, if $e\otimes \omega \in V\otimes V^*$, then it defines a map
$$
v \mapsto \langle \omega,v\rangle e.
$$
In particular, given any frame of vector fields, $(\vec{e}_1, \dots, \vec{e}_n)$ with the dual basis of $1$-form $(\omega^1, \dots, \omega^n)$, the definition of $d\vec{r}$ is
$$
d\vec{r} = \vec{e}_i\otimes \omega^i
$$
This definition is invariant under change of basis. In particular, the map associated with it is simply the identity map,
$$
\langle d\vec{r}, v\rangle = \langle \vec{e}_i\otimes \omega^i, v^je_j\rangle
= \vec{e}_iv^j\langle \omega^i,e_j\rangle = v^ie_i = v.
$$
If you have a coordinate system $(x^1, \dots, x^n)$, then you can set $(\vec{e}_1, \dots, \vec{e}_n)$ equal to the coordinate vector fields and $(\omega^1, \dots, \omega^n) = (dx^1, \dots, dx^n)$.
If you have another coordinate system $(y^1, \dots, y^n)$ with corresponding coordinate vector fields $(\vec{f}_1, \dots, \vec{f}_n)$ and dual frame $(dy^1, \dots, dy^n)$, then
$$
d\vec{r} = \vec{e}_i\otimes dx^i = \vec{f}_i\otimes dy^i.
$$
From here, I think it should be straightforward to derive all of the formulas you've written down.
