Semantics: basis of tangent vectors in Rn vs basis of position vectors in Rn The tangent plane of any point in Rn (usual structure) is linearly isomorphic to Rn itself. What is clear is that position vectors and tangent vectors are not the same kind of object even in Rn, as demonstrated by curvilinear coordinate transforms. Yet we still use the same ol' unit vectors to describe them both (e1, e2, e3 or i, j, k etc). 
It strikes me as problematic that position vectors "look" the same as tangent vectors. On top of that, the basis position vectors are easy to visualize as unit displacements in a reference direction, but the basis tangent vectors (in general) aren't even geometric vectors but differential operators (e.g., the basis of partials). So my question is this: how does one define the position vector basis (when it exists) as distinct from the tangent vector basis?
I thought a bunch on it on my own before posting here, but I came up with nothing conclusive. 
My first reaction was that it doesn't make sense to write points as vectors in a curvilinear coordinate system. Algebraically, position vectors don't add like vectors in a linear space unless you treat the curvilinear basis vectors as fields. But if you work carefully over the whole tangent bundle, you can get away with using position vectors with the same basis as the tangent space. Now, if tangent vectors are supposed to behave like differential operators, then a linear combination of the basis tangent vectors should give you another operator, not a point. So what's going on? How did you get from a differential operator to a point?
My best guess was that the projection map was the key. That seems to work, since you shouldn't be able to pull the same trick on the Möbius strip or something non-orientable. Am I on the right track? I'm still just beginning differential geometry, so the idea of the projection map isn't very clear to me. 
 A: Clarification: For any $n$-dimensional manifold $M$ and $p\in M$, the tangent space $T_pM$ is isomorphic to $\mathbb{R}^n$. However, it is canonically isomorphic to $\mathbb{R}^n$ only when $M$ is an open subset of $\mathbb{R}^n$.
It is true that often the same notation is used for elements in $\mathbb{R}^n$ and tangent vectors to a point. This is probably the result of some confusion. However, the "correct" notation does differentiate between the two. Points in Euclidean space are usually denoted by either row or column vectors. Tangent vectors can be denoted by linear combinations of $\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_n}.$ 
Another clarification: Tangent vectors can (and need to) be visualized as geometric directions. The tangent vector $\frac{\partial}{\partial x_1}$ at a point $p$ is the set of paths $$\{\gamma:(-\epsilon,\epsilon)\to\mathbb{R}^n|\gamma(0)=p,\dot{\gamma}(0)=(1,0,\ldots,0)\}.$$ It acts as a derivation in the following manner: given a function $f$, we define$$\left.\frac{\partial }{\partial x_1}\right|_p(f)=\left.\frac{d}{dt}\right|_{t=0}f\circ\gamma,$$where $\gamma$ is any path in the above set.
