Is it "mathematically wrong" to ignore dual spaces, 1-forms, and covariant/contravariant indices in classical mechanics? If everything you are working with is in Euclidean 3-space (or $n$-space) equipped with the dot product, is there any reason to bother with distinguishing between 1-forms and vectors? or between covariant and contravariant tensor components?  I'm fairly certain that if you do not, then none of your calculations or relations will be numerically wrong, but are they "mathematically wrong"?
Example: I'll write some basic tensor relations without distinguishing between vectors and 1-forms/dual vectors or between covariant/contravariant components. All indices will be subscripts. Tell me if any of the following is wrong:
I would often describe a rigid  transformation of an orthonormal basis (in euclidean 3-space), ${\hat{\mathbf{e}}_i}$, to some new orthonormal basis ${\hat{\mathbf{e}}'_i}$ as
$$
\hat{\mathbf{e}}'_i\;=\; \mathbf{R}\cdot\hat{\mathbf{e}}_i \;=\; R_{ji}\hat{\mathbf{e}}_j
\qquad\qquad (i=1,2,3)
$$
For some proper orthogonal 2-tensor ${\mathbf{R}\in SO(3)}$  (or whatever the tensor equivalent of $SO(3)$ is, if that's a thing). It's then pretty straightforward to show that the components, $R_{ij}$, of ${\mathbf{R}}$ are the same in both bases and $\mathbf{R}$ itself is given in terms of ${\hat{\mathbf{e}}_i}$ and ${\hat{\mathbf{e}}'_i}$ by
$$
\mathbf{R}=R_{ij}\hat{\mathbf{e}}_i\otimes\hat{\mathbf{e}}_j = R_{ij}\hat{\mathbf{e}}'_i\otimes\hat{\mathbf{e}}'_j = \hat{\mathbf{e}}'_i\otimes\hat{\mathbf{e}}_i
\qquad,\qquad\quad
R_{ij}=R'_{ij}=\hat{\mathbf{e}}_i\cdot\hat{\mathbf{e}}'_j
$$
Then, given the basis transformation in the first equation, the components of some vector $\vec{\mathbf{u}}=u_i\hat{\mathbf{e}}_i=u'_i\hat{\mathbf{e}}'_i$ and some 2-tensor $\mathbf{T}=T_{ij}\hat{\mathbf{e}}_i\otimes \hat{\mathbf{e}}_j = T'_{ij}\hat{\mathbf{e}}'_i\otimes \hat{\mathbf{e}}'_j$
would transform as
$$
u'_i = R_{ji}u_i \qquad \text{matrix form: } \qquad [u]'= [R]^{\top}[u]
\\
T'_{ij} = R_{ki}R_{sj}T_{sj}  \qquad \text{matrix form: } \qquad
[T]' = [R]^{\top}[T][R]
$$
and for some p-tensor we would have
$$S'_{j_1j_2\dots j_p} \;=\; \big(
     R_{ i_1j_1}R_{ i_2j_2} \dots R_{ i_pj_p}
     \big) S_{ i_1 i_2\dots i_p}
$$
and if ${\hat{\mathbf{e}}_i}$ is an inertial basis and  ${\hat{\mathbf{e}}'_i}$ is some rotating basis, then  the skew-symmetric angular velocity 2-tensor of the ${\hat{\mathbf{e}}'_i}$ basis realtive to  ${\hat{\mathbf{e}}_i}$ is given by
$$
\boldsymbol{\Omega} \;=\; \dot{\mathbf{R}}\cdot\mathbf{R}^{\top}
\qquad,\qquad
\text{componenets in } \hat{\mathbf{e}}_i \;:
\qquad  \Omega_{ij} = \dot{R}_{ik}R_{jk}
$$
Or, in matrix form (in the $\hat{\mathbf{e}}_i$ basis) the above would be $[\Omega]=[\dot{R}][R]^{\top}$. The third equation can be used to convert to the $\hat{\mathbf{e}}'_i$ basis.  The familiar angular velocity (pseudo)vector is then given by
$$
\vec{\boldsymbol{\omega}}= -\tfrac{1}{2}\epsilon_{ijk}(\hat{\mathbf{e}}_j\cdot \boldsymbol{\Omega}\cdot \hat{\mathbf{e}}_k)\hat{\mathbf{e}}_i
\qquad,\qquad
\text{componenets in } \hat{\mathbf{e}}_i \;:
\qquad
\omega_i = -\tfrac{1}{2}\epsilon_{ijk}\Omega_{jk}
$$
where $\epsilon_{ijk}$ are the components of the Levi-Civita 3-(pseudo)tensor, $\pmb{\epsilon}$, which itself may be written in any right-handed orthonormal bases as
$$
\pmb{\epsilon} = \epsilon_{ijk}\hat{\mathbf{e}}_i\otimes\hat{\mathbf{e}}_j \otimes \hat{\mathbf{e}}_k = \tfrac{1}{3!}\epsilon_{ijk}\hat{\mathbf{e}}_i\wedge\hat{\mathbf{e}}_j \wedge \hat{\mathbf{e}}_k
= \hat{\mathbf{e}}_1\wedge\hat{\mathbf{e}}_2 \wedge \hat{\mathbf{e}}_3
\quad,\quad \epsilon_{123}=1
$$
The time-derivative of some vector $\vec{\mathbf{u}}=u_i\hat{\mathbf{e}}_i=u'_i\hat{\mathbf{e}}'_i$ would then be given in terms of the components in the inertial and rotating bases by the familiar kinematic transport equation
$$
\dot{\vec{\mathbf{u}}} = \dot{u}_i\hat{\mathbf{e}}_i = \dot{u}'_i\hat{\mathbf{e}}'_i + \boldsymbol{\Omega}\cdot\vec{\mathbf{u}} \;=\; (\dot{u}'_i + \Omega'_{ij}u'_j  )\hat{\mathbf{e}}'_i
$$
where $\boldsymbol{\Omega}\cdot\vec{\mathbf{u}} = \vec{\boldsymbol{\omega}}\times\vec{\mathbf{u}}$.
end example
question: So, I'm pretty sure that none of the above would give me numerically incorrect relations. But I called everything either a vector, 2-tensor, or 3-tensor. Nothing about forms, (1,1)-tensors, (0,2)-tensors, dual vectors, etc.  Is the above formulation mathematically ''improper''? For instance, do I need to write ${\mathbf{R}}$ as a (1,1)-tensor, ${\mathbf{R}}=R^{i}_{\,j}\hat{\mathbf{e}}_i\otimes\hat{\boldsymbol{\sigma}}^j$, using the basis 1-forms $\hat{\boldsymbol{\sigma}}^j$?  Does the angular velocity tensor need to be written as a 2-form or (0,2)-tensor?
context:  My BS is in physics and I am currently a PhD student in engineering. Aside from a graduate relativity course I took in the physics department, I have never once seen  raised indices or mention of dual vectors/1-forms in any class I have ever taken or in any academic paper, I have ever read. That was until I recently started teaching myself some differential geometry in hopes of eventually understanding Hamiltonian mechanics from the geometric view. So far, I have mostly only succeeded in destroying my confidence in my knowledge of basic tensor algebra involved in classical dynamics.
 A: No, it's not mathematically wrong. To show it mathematically wrong, you'd need to show that it leads to a contradiction.
As for what you should use in physics, that depends on the problem, how your intuition works, how it helps with the formulation, whether it illuminates or obfuscates what's going on, ...
A: As long as you restrict yourself to orthonormal bases, then that's fine.  The reason for this is that indices are "raised" or "lowered" via the metric, and in an orthonormal basis the metric components are $g_{ij}=\delta_{ij}$.
As soon as your basis is non-orthonormal, however, this goes out the window.  There are many good reasons to use non-orthonormal bases in various circumstances, but since you've explicitly stated that you'd ultimately like to understand Hamiltonian mechanics from a geometrical standpoint, I'll highlight the most glaring problem: in Hamiltonian mechanics on a symplectic manifold, there is no metric, and so the entire concept of orthonormality goes out the window.
It is still useful to define an isomorphism between tangent vectors and their duals on a symplectic manifold, but we need something other than a metric to do so. The structure we use is the symplectic form $\Omega$, which is by definition antisymmetric; this immediately implies that $\Omega_{ij}=\delta_{ij}$ is ruled out as a possibility in any coordinate system.  As a result, vectors and their duals always have different components, and distinguishing between them and their transformation behaviors is crucial.
A: It is very much wrong in my opinion. If you don't the contravariant, co-variant distinction, then the transformation rule of different derivative quantities like the gradient, curl, div etc doesn't make sense. Many basic Vector calculus books derive for each coordinate system through geometry, and tells to find the conversion rules between coordinate system using geometry
But, it maybe unclear to a student how it all algebraically relates. To get the right algebraic conversion rules of the expressions between coordinate systems, we can not do it without Covariant and contravariant indices.
