My biggest issue is with the coordinate-definition of tensor products. A physicist defines a rank $k$ tensor over a vector space $V$ of dimension $n$ to be an array of $n^k$ scalars associated to each basis of $V$ which satisfy certain transformation rules; in particular, if we know the array for a given basis, we can automatically determine it for a different basis. Another way to say this is that the space of tensors is the set of pairs consisting of a basis and a $n^k$ array of scalars, identified by an equivalence relation which gives the coordinate transformation law. For some strange reason, people seem to call this a coordinate-free definition. While it is in a sense coordinate-free (the transformation between coordinates lets you break free of coordinates in a sense), it is very confusing at first sight. People who use this definition will they say that certain operations are coordinate-free. What they mean by this, and it took me a long time to figure this out, is that you can do a certain algebraic operation to the coordinates of the tensor, and the formula is the same no matter which basis you work with (e.g., multiplying a covariant rank $1$ tensor with a contravariant rank $1$ tensor to get a scalar, or exterior differentiation of differential forms, or multiplying two vectors to get a rank $2$ tensor).
The much nicer definition uses tensor products. This is a coordinate-free construction, as opposed to the coordinate-full description given above. This definition is nice because it connects to multilinear maps (in particular, it has a nice universal property). It also helped me see why tensors are different from elements of some $n^k$-dimensional vector space over the same field (they are special because we are equipped not just with a vector space but with a multilinear map from $V \times \cdots \times V \to V \otimes \cdots \otimes V$. The covariant/contravariant distinction can be explained in terms of functionals. This allows you to talk about contraction of tensors without worrying having to prove that it is coordinate-invariant! Finally, once you have all that under your best, you can easily derive the coordinate transformation laws from the multilinearity of $\otimes$.
