Hessian as a tensor, multi-dimensional taylor series, and generalizations

No, no, no! You left out a term involving $\frac{df(y(x))}{dy}\frac{d^2y}{dx^2}$. This term vanishes at critical points -- points where $df=0$ -- so that indeed at such a point the Hessian define a tensor -- a symmetric bilinear form on the tangent space at that point -- independent of coordinates. Paying attention to what kind of bilinear form it happens to be is the beginning of Morse theory, but it's only intrinsically defined as a tensor if you're at a critical point.

Notice that even the question of whether the Hessian is zero or not is dependent on coordinates. Even in a one-dimensional manifold.

Taylor polynomials don't live in tensor bundles, but in something more subtle called jet bundles.

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.

More generally, for any vector bundle $V \to M$, the vanishing of a section at a point to a given order $k$ is independent of local coordinates and choice of basis of local sections, and when this happens at some point $m \in M$, the $(k+1)$-jet of the local section at that point is a well defined element of $S^kT^*_mM \otimes V_m$.