Why do we use $\partial_\mu$ to contract the inverse vielbein, but $dx^\mu$ to contract vielbein? I'm studying the topic "Lie group manifold". Previously, I have been exposed to Vielbein $e_\mu^m$ and its inverse $e_m^\mu$ in the course of General Relativity, where those are just numbers, no problem.
However, this is my first time seeing people contracting $e_m:=e^\mu_m \partial_\mu$ and $e^m:=e_\mu^m dx^\mu$, which is not quite "symmetric" to me, because we can certainly do $[e_m,e_n]$ but not $[e^m,e^n]$ (since the first one contains derivative, but not the later).
I want to know why people treat, more generally, covariant vectors and contravariant vectors differently? One basically becomes an operator, the other is "numerically" infinitesimal; and what's the potential purpose of doing this?

Forgive me if this is something really basic in differential geometry, I didn't receive a holistic education on that as a physics graduate student.
 A: The modern way is to construct objects that are intrinsically defined without reference to coordinates.
On a smooth manifold the only naturally existing "basic" objects are smooth functions (and maybe smooth curves), so all other objects must be defined from smooth functions and curves. Actually, functions are enough.
Let $C(M)$ be the commutative ring/algebra of smooth functions on the (smooth) manifold $M$.
Define for any $x\in M$ the space $T_xM$ of point-derivations of $C(M)$. Eg.  $v\in T_xM$ is an $\mathbb R$-linear map $v:C(M)\rightarrow\mathbb R$ that satisfies for $f,g\in C(M)$ the "pointed Leibniz rule" $$ v(fg)=v(f)g(x)+f(x)v(g). $$
Further analysis reveals that this space is finite dimensional, and if $(U,\mathbf x)$ is a local coordinate system of $M$ with $x\in U$, then the elements $\partial_\mu |_x$ which act on a function by $$ \partial_\mu|_x(f)=\frac{\partial (f\circ\mathbf x^{-1})}{\partial x^\mu}(\mathbf x(x)) $$ form a linear basis for $T_xM$, so any element of $T_xM$ can be written as $$ v=v^\mu\partial_\mu|_x $$for some numbers $v^\mu$.
The set of all tangent vectors at $x\in M$ is identified with the just-constructed space $T_xM$. There is an intuitive justification for that, namely, a tangent vector should represent a "velocity-type" object, and velocity can be characterized how arbitrary functions are changing as we move.

Covectors are dual objects to vectors. From this point on, they can be constructed by duality, eg. we take $T_x^\ast M$ to be the set of all linear maps $\omega: T_xM\rightarrow \mathbb R$, eg. that associate to any tangent vector $v\in T_xM$ a number $\omega(v)$ in a linear fashion.
But covectors can also be constructed explicitly from functions. If $\mathcal I_x<C(M)$ is the set of all smooth functions that vanish at $x\in M$, then $$ T^\ast_xM\cong \mathcal I_x/\mathcal I_x^2. $$ The heuristic justification for this procedure can be found in my physics stack exchange answer here.
If $f\in C(M)$ is an arbitrary smooth function, then we can define the differential of $f$ at $x$ as $$ \mathrm df|_x\in T^\ast_xM,\ \mathrm df|_x(v)=v(f),\ \forall v\in T_xM. $$
This definition is far more natural from the point of view from the $\mathcal I_x/\mathcal I_x^2$ construction, where $d|_x$ is simply the natural projection from $C(M)$ to $\mathcal I_x/\mathcal I^2_x$, eg. $\mathrm df|_x=[f-f(x)]$.
As it turns out, if $(U,\mathbf x)$ is a coordinate system, then $\mathrm dx^\mu|_x$ is a basis for $T^\ast_xM$, and any $\omega\in T^\ast_xM$ can be expressed as $$ \omega=\omega_\mu\mathrm dx^\mu|_x $$ for some numbers $\omega_\mu$.
The basis obtained this way is dual to the $\partial_\mu|_x$ basis in the sense that $$ \mathrm dx^\mu|_x(\partial_\nu|_x)=\partial_\nu|_x(x^\mu)=\delta^\mu_\nu. $$

If we consider fields, then we can take $\partial_\mu$ to be local vector fields defined in a coordinate domain, and $\mathrm dx^\mu$ to be local covector fields defined in the same coordinate domain and expand fields about them too.
However not every field of basis vectors is of the form $\partial_\mu$ for some coordinate system, likewise not every field of basis covectors is of the form $\mathrm dx^\mu$ for some coordinate functions $x^\mu$.
One can show, that if $e_a$ (I intentionally use a different index here) are $n$ ($n=\dim M$) pointwise linearly independent local vector fields, then there is a coordinate system $u^a$ such that $e_a=\partial_a$ if and only if the local vector fields $e_a$ commute, eg. $$ [e_a,e_b]=C^c_{ab}e_c=0. $$
(If this happens, they are called holonomic)
From the dual perspective, if $e^a$ are the local basis field of covectors dual to $e_a$, eg. $e^a(e_b)=\delta^a_b$
, then $$ \mathrm de^a=-\frac{1}{2}C^a_{bc}e^b\wedge e^c $$ (some knowledge of differential forms is needed to understand this),
 thus $e^a=\mathrm du^a$ for some $u^a$ only if $\mathrm de^a=0$.

In OPs case, they are given a coordinate system $(U,\mathbf x)$ with coordinate basis $\partial_\mu$ and dual coordinate basis $\mathrm dx^\mu$, and they are also given a possibly nonholonomic basis field $e_a$ (for simplicity, assume that they are also defined on $U$), and corresponding dual basis fields $e^a$.
Then because both sets are bases, there are locally defined $\mathrm{GL}(n,\mathbb R)$-valued functions $e^\mu_a$ and $e^a_\mu$ (pointwise matrix inverses of one another) such that $$ e_a=e^\mu_a\partial_\mu \\ \partial_\mu=e^a_\mu e_a \\ e^a=e^a_\mu\mathrm dx^\mu \\ \mathrm dx^\mu =e^\mu_a e^a. $$
These relations should be interpreted as expressing one sets of basis (co)vectors with respect to another.
As it is usual in physics literature and older mathematics literature as well, the "invariant objects" $\partial_\mu,e_a,\mathrm dx^\mu,e^a$ are sometimes omitted, and only the components $e^\mu_a,e^a_\mu$ are manipulated.
In the approach preferred by this answer, the functions $e^\mu_a$ and $e^a_\mu$ however serve only the role of expansion coefficients associated to the intrinsically defined objects.
