It is important in differential geometry that the space of derivations at a point $p\in V$ (where $V$ is a finite-dimensional real vector space) is isomorphic to the space of partial derivatives (or velocity vectors) at $p$. This can be proven using a very weak version of Taylor's theorem, which is the approach taken in Tu's "Intro to Manifolds" and described here.
Meanwhile it has also been described in theseposts how Taylor's theorem can be stated in a completely coordinate-free form. Therefore, it is natural to suspect that Tu's proof can also be rewritten in a coordinate-free form. However, I can't see how to actually carry this out.
Tu's proof uses the following version of Taylor's theorem.
$$f(x) = f(p) + \sum (x^i - p^i)g_i(x) \qquad g_i(p) = \frac{\partial f}{\partial x^i}(p)$$
Meanwhile the coordinate-free version of the same theorem would look something like
$$f(x) = f(p) + \tilde{g}(x - p), \qquad \tilde{g}(x - p) = o(\|x - p\|).$$
Tu's approach seems to rely explicitly on the ability to express $\tilde{g}$ as a sum of products $(x^i - p^i)g_i(x),$ and I can't figure out whether this idea (and the remainder of the proof) can be expressed without coordinates.