> But I would go with something more like: tensors are a way to represent vectors so that the representation of a given vector is the same no matter what basis (or coordinate system) you choose for your vector space.

That's just incorrect though for a couple of reasons. Firstly, a vector in the sense in which it is used in physics is a rank 1 tensor so it has this transformation behaviour just like other higher order tensors. Secondly the representation is the thing that changes, but the meaning of that representation in the old basis and the new basis is the same. For example, if I take the displacement from me to the top of the Eiffel tower, I can represent that in xyz Cartesian coordinates or in spherical or cylindrical coordinates, or I can measure it relative to an origin that starts with me or at sea level at 0 latlong. The representation will be very different in each case, but the actual displacement from me to the top of the Eiffel tower doesn't change. What has happened is the basis vectors transform in exactly such a way as to make that happen. It's a rank 1 tensor in 3 dimensions because there is a magnitude and one direction (one set of 3 basis vectors) in whatever case.

Now if I want an example of a rank 2 tensor think about a stress tensor. I have a steel beam which is clamped at both ends and a weight is on top of it. This is a tensor field. For every point in the beam there are different forces acting in each direction. So you could imagine the beam as made up of a grid of little rubik's cubes. On each face of each cube you have different net forces. (eg at the middle of the beam the forces are mainly downwards due to gravity, at the ends of the beam the fact that the middle of the beam is bowing downards will lead to the "faces" that point to the middle of the beam to be being pulled towards the middle (transverse to the beam and slightly downwards) whereas the opposite face is pulled in the opposite direction because the ends of the beam are clamped. So I need two sets of basis vectors. One set indicates the "face" experiencing the force, one set indicates the direction of the force. Now just like the vector/rank one tensor case I can represent those in whatever coordinate system I want, and my representation will be different in each case, but will mean the same sets of forces in the same directions and applied to the same directions because both sets of basis vectors will transform to make that true. I would call that a rank 2 tensor field because I would express it as a function from a set of spatial coordinates to a thing which has a magnitude and 2 directions (that's what I think of as the tensor). However I understand physicists and civil engineers and stuff just call the whole thing the stress tensor (not the stress tensor field). I could be wrong.

So what I mean when I talk about the reality of the tensor I mean whatever it is the tensor is expressing in the physical universe (eg the displacement from me to the tower or the stress in the beam). From a mathematical point of view I agree of course, mathematical objects themselves are purely arbitrary and abstract. But if you have a bridge and you want to make sure it doesn't buckle and fall down, the stress tensor in the bridge is a real and important fact of the universe that you need to have a decent understanding of.

> That's just incorrect though

Quite possible. But that's in no small measure because I have yet to find an actual cogent definition of "tensor" that distinguishes a tensor from an array. (I have a similar problem with monads.)

> So what I mean when I talk about the reality of the tensor I mean whatever it is the tensor is expressing in the physical universe

OK, but then "the reality of a tensor" not depending on the coordinate system has nothing to do with tensors, and becomes a vacuous observation. It is simply a fact that actual physical quantities don't depend on how you write them down, and hence don't change when you write them down in different ways.