Upper and lower indices when identifying tangent space vectors with underlying vector space Given an $n$-dimensional manifold $M$, we can construct a basis for each tangent space $T_pM$ from a local coordinate basis $\{x^i\} \subset \mathbb{R}^n$ as
$$\vec{e}_i = \left( \frac{\partial}{\partial x^i} \right) \quad \forall\, i \in \{1,2,\ldots,n\}$$
(as is done e.g. here). It is then conventional to denote vectors $\vec{v} \in T_p M$ in this basis as $\vec{v} = v^i \vec{e}_i$, where we can interpret the raised index in $v^i$ as a sign that the vector is contravariant, or that it is a $(1,0)$ tensor that assigns to the tangent basis vector $\vec{e}_i$ a value $\in \mathbb{R}$.
Now, if $M$ is a vector space $V$, we can identify the tangent vectors $\vec{v} \in T_p V$ at any point $p \in V$ with vectors from $V$ itself. My main question is whether there is any justification for, or anything deep that can be understood from, the fact that when we perform this identification, the indices are somehow not where they should be. If we take the local coordinate basis $x^i$ from before as a global basis in $V$, then it seems that:
$$\vec{v} = v^i \vec{e}_i \in T_p V \quad \text{can be identified with} \quad \vec{v}' = v^i x^i \in V \quad\text{(or should it be }v_i x^i \in V\text?).$$
Since the position of the index is significant when applying tensors to these quantities, I want to make sure I don't just gloss over this as just an artifact of the notation. Specifically, the position of the index in the tangent space basis vector $\vec{e}_i$ seems to clash with the corresponding vector space basis component $x^i$ when we identify one as the other.
Is the answer just to always cleanly separate between the elements of $T_pV$ and $V$ itself, even if there is a bijection between them?
Context: Took a mathematical introduction to manifolds and am trying to reconcile the concepts and notation used in (special) relativity, where the distinction between $M$ and its tangent spaces is routinely dropped, with the notions from manifold theory.
 A: If $\{\xi_1, \dots, \xi_n\} \subset V$ is a basis for a vector space $V$, then it gives rise to a dual basis $\{x^1, \dots x^n\}$ for the dual space. i.e $x^i:V \to \Bbb{R}$ are linear functions, which means each $x^i$ is an element of $V^*$, NOT $V$. So, given any vector $\vec{v} \in V$, we can always write it as a linear combination of $\{\xi_1, \dots, \xi_n\}$ as
\begin{align}
\vec{v} &= \sum_{i=1}^n v^i \xi_i
\end{align}
and by unwinding the definition of the dual basis, you can check for yourself that $v^i = x^i(v)$, i.e the evaluation of $x^i:V \to \Bbb{R}$ on the vector $\vec{v} \in V$. So, using the summation convention,
\begin{align}
\vec{v} = v^i \xi_i = x^i(v)\cdot \xi_i \in V
\end{align}
You're probably getting confused because of your own terminology; for example in the very first line you write the misleading statement:

... can construct a basis for each tangent space $T_pM$ from a local coordinate basis $\{x^i\}\subset \Bbb{R}^n$ as...

It is not true that $x^i \in \Bbb{R}^n$, so it is definitely not true that $\{x^i\}$ forms a "local coordinate basis". What it is, is a local coordinate system/chart. When we intuitively speak of coordinates, what we think of is that every point has associated to it a certain set of numbers; so if we translate this directly into more formal language, what it amounts to is $n$-functions $x^i: M \to \Bbb{R}$.

Note that it is perfectly fine to identify $T_pV$ with $V$, because you get this isomorphism simply from the existing data of the smooth structure (i.e if $(M=V, \mathcal{A})$ is your smooth manifold and $\mathcal{A}$ is the maximal smooth atlas containing the identity chart $(V,\text{id}_V)$, then this automatically induces an isomorphism of each tangent space with the underlying vector space).
However, what is not fine is to arbitrarily invoke isomorphisms of $V$ with $V^*$, without any further structure (like a pseudo-inner product). Also, in some cases, for example if you have a basis $\{\xi_1, \dots, \xi_n\}$ of $V$ and the dual basis $\{x^1, \dots, x^n\}$ of $V^*$, then yes, you can get an isomorphism $V \to V^*$. But, just because you can do so, doesn't mean you should (I for one never like to mix up my vector space with its dual, because it is very easy to quickly get confused about what is actually going on).
