Is it ever possible to fully avoid Cartesian coordinates? I think this question is more philosophical than mathematical, though I may be wrong, probably it is just a stupid lack of understanding. Anyway, if you don't mind, please read the question carefully to get the sense of what I am asking. It is a bit difficult to express.
Tensors and differential geometry consider oblique (non-orthogonal) and even curvilinear "axes" or basis vectors.  But, the basis vectors themselves must be expressed somehow.  Does this require going back to some underlying "master" orthogonal (Cartesian) coordinate system?
Here are two examples of my puzzle.
Example 1. 
A vector $\mathbf{v}$ can be written in terms of Cartesian coordinates as
$$
   \mathbf{v} = v^k \mathbf{e}_k
$$
where $e_k$ are the regular basis vectors (in the 3D case, (1,0,0), (0,1,0), (0,0,1)).
A vector can also be written in terms of an arbitrary non-orthogonal system as for example
$$
   \mathbf{v} = w^k \mathbf{u}_k
$$
Suppose that $w^k$ are given. To know the vector, we have to know what the vectors $\mathbf{u}_k$ are, and these would either have to be specified in terms of an underlying Cartesian coordinate system, for example
$$
   \mathbf{u}_2 = (0.3, 0.799, -0.1)
$$
or in terms of some other coordinate system, which in turn(!) (possibility of infinite regress here) would need to be specified eventually in terms of the Cartesian coordinate system?
I suppose if the other coordinate system is a physical given thing (lines drawn by aliens in the desert)
then one can avoid the issue.
Example 2.
In Susskind's Einstein's General Theory of Relativity | Lecture 5
https://www.youtube.com/watch?v=WtPtxz3ef8U
at around 18:30 he describes why the derivative of the components of a tensor are not themselves a tensor, giving the example of a field of constant-direction-and-constant-magnitude vectors. The derivatives of components of these vectors w.r.t. cartesian axes are zero, but the derivatives w.r.t spatially varying axes are non-zero.
Here we have some vectors, which appear to be pointing in the same direction, and measure them using some underlying axes, which certainly appear to be curving.  But how do we know that the vectors are pointing in the same direction and the axes are curving and not the other way?  In everyday live, of course I can see that what was drawn on the white board was straight/curving, but that is not a scientific answer. Don't we need to measure with respect to some other coordinate system,  and wouldn't that one be...(eventually)...Cartesian?
Note, it sounds that someone in the class was possibly asking this question around 18:30, but the professor closed the question.
EDIT: Seeing the first 3 answers, it is clear that I failed to explain the question... or else the question is just nonsense. I fear that I am missing something very fundamental! Surprising however. The idea of a basis is something I have successfully used in various exercises and is second nature at this point.
 A: I think there is a fine line between what is a coordinate-free treatment of concepts (which I think is mostly what you're after), and precision. In my opinion, coordinate-free treatments are excellent for understanding and fitting together high-level concepts, while being rather cumbersome for explicit calculations, while coordinate-based computations are often much faster and easier.
For your first example, a specific vector is annoying to pin down without coordinates in, for example, $\mathbb{R}^2$. This is due to there not being any extra structure on the space. However, linear maps on the space can often be given unambiguously without coordinates, for example


*

*The identity map.

*The zero map.

*The map scaling every vector up by 2.

*The map rotating by $\pi/2$ anticlockwise.


We can make some explicit computations with some of these, for example that last map $r$ satisfies $r^2 = -1$, and so its characteristic polynomial must be $r^2 + 1$, from which we can see it has no real eigenvalues. At no point have I needed to use any basis here.
For your second example, note that vectors are always straight - this is part of what it is to be a vector: it behaves like a "straight" thing under addition and scalar multiplication. If I take a vector at a point in any coordinate system (even a curved one), and scale it by $2$, that will mean its coordinates are scaled by $2$. The frame (collection of coordinate vectors) may change from point to point, but stays constant on a point. The tangent bundle is a good example of this: your space is curved, but vectors you consider at a point are trapped inside a well-behaved vector space.
Finally, differential geometry is done on a real or complex manifold, where each patch of space looks locally like $\mathbb{R}^n$. So the definition of a manifold has Euclidean space "baked-in", so you will always be able to express things as eventually cartesian coordinates if you take small enough patches. But that does not mean the entire manifold looks like Cartesian space.
A: It appears you're trying to get at the distinction between differential topology and differential geometry.
In each field, an $n$-manifold is a topological space locally modeled on open subsets of Cartesian $n$-space. In that sense, there is an underlying "standard model", and computations involving tensor fields (functions, vector fields, differential forms, Riemannian metrics, etc.) can be expressed in local coordinates, which we may as well take to be standard Cartesian coordinates.
The subtle distinction is, in differential topology the only meaningful concepts are those independent of the coordinate system. This limits objects of consideration to the first derivative (the tangent bundle, vector fields) and exterior calculus (differential forms, the exterior derivative, and integration of forms). If memory serves, Spivak's Comprehensive Introduction to Differential Geometry, Volume I discusses the technicalities in readable detail. (Fun fact: On a smooth manifold, a smooth function has no invariant second derivative. There is, however, a meaningful notion of index of the Hessian at a critical point, which is all that Morse theory requires.)
In differential geometry, by contrast, one fixes a metric and correspondingly restricts the notion of differentiation by requiring suitable compatibility with the metric. The resulting covariant differentiation makes sense for arbitrary tensor fields.
I didn't watch the linked Susskind lecture, but covariant differentiation can meaningfully distinguish between "straight vectors, curving axes" and "curving vectors, straight axes". This is part of what is meant by "the metric determines the geometry": a physical concept such as "straight" doesn't make sense in a smooth manifold, only with respect to a metric.
(At risk of self-promotion, related issues are addressed in my answer to What does the integral of position with respect to time mean?.)
A: You write:
or in terms of some other coordinate system, which in turn(!) (possibility of
infinite regress here) would need to be specified eventually in terms of the
Cartesian coordinate system?

This is actually not required.  Consider your first example:
A vector v can be written in terms of Cartesian coordinates as v=$v^k$ $e_k$
But then what is $e_k$ to be written in terms of?  Answer: NOTHING - this is taken as a primitive.  Likewise, you don't NEED to specify the given coordinate system in terms of another; you can simply take its elements as primitives, and reason directly using those primitives.  Of course, as others say, from a pragmatic perspective, certain things may be easier to calculate in a different coordinate system.
