Sorry for reviving this question. Everything Tom said is correct, but there is more to say about "coordinate-free Taylor series".

It is true that arbitrary jet bundles $J^k(M,N)$ are subtle. They are affine bundles, but in general, they are not vector bundles. The special case $J^k(M,\mathbb{R})$ is a vector bundle and this is where the Taylor series of a map $f : M \to \mathbb{R}$ lives. The vector bundle $J^k(M,\mathbb{R})$ is not a tensor power of the tangent bundle as Tom pointed out, but it is far easier to understand than an arbitrary jet bundle. Indeed, the structure group of $J^k(M,\mathbb{R})$ is sometimes called the Phylon group.

Sorry for reviving this question. Everything Tom said is correct, but there is more to say about "coordinate-free Taylor series".

It is true that arbitrary jet bundles $J^k(M,N)$ are subtle. The fibers are vector spaces but the transition maps are not linear. The special case $J^k(M,\mathbb{R})$ is a vector bundle and this is where the Taylor series of a map $f : M \to \mathbb{R}$ lives. The vector bundle $J^k(M,\mathbb{R})$ is not a tensor power of the tangent bundle as Tom pointed out, but it is far easier to understand than an arbitrary jet bundle. Indeed, the structure group of $J^k(M,\mathbb{R})$ is sometimes called the Phylon group.