Jet bundle - affine bundle Suppose $X\rightarrow V$ is a fibre bundle. $X^{(r)}$ denotes the bundle whose fibre at each $x$ is the set of all equivalence classes of $r$-jets.
Since $r$-tangency is a finer equivalence relation for larger $r$, it follows that there are canonical projections $X^{(r)}\rightarrow X^{(r-1)}$.
An exercise says that this is a bundle with affine fibre. I do know, a space being merely affine implies, that there's no canonical choice of a zero vector but that there's a transitive action of a vector space on it.
But the projection is just $(x,f(x),...D^kf(x) ... D^{r-1}f(x),D^rf(x))\rightarrow (x,f(x),...D^kf(x) ... D^{r-1}f(x))$
where $D^kf(x)$ is the tuple of all order $k$ partial derivatives of $f$ at $x$.
I don't understand why only 'affine' and why isn't there a canonical zero section. Is it because you can't always find $g$ which is ($r-1$)-tangent to $f$ at $x$ and has all order $r$ derivatives vanishing at $x$?
Thankful in advance for clarification.
 A: Let me give a simple example. 
Let your manifold be $\mathbb{R}$. We consider two different parametrisations (coordinates) satisfying 
$$ y = x + x^3 $$
(Note that $x\mapsto y$ is bijective, with derivative $\partial_x y = 1 + x^2 \geq 1$ and so is a diffeomorphism of $\mathbb{R}$ to itself.)
Let $f(x)$ be the function $x + x^3$. In terms of the coordinates $y$ you have $f(y) = y$. 
In terms of the $y$ coordinates, the higher derivatives $f^{(k;y)}(y) = 0$ for all $k > 1$. But in terms of the $x$ coordinates, you have that, in particular, $f^{(3;x)}(x) \equiv 6 \neq 0$. 
Remember that jets is supposed to be a coordinate invariant way of looking at the "Taylor series" associated to a function, and we see here that whether the $k$-th derivative of a given function vanishes can depend on the coordinates chosen. 
So, a propos

Is it because you can't always find $g$ which is $(r−1)$-tangent to $f$ at $x$ and has all order $r$ derivatives vanishing at $x$?

Your interpretation is not strictly correct. Fix a coordinate system in a neighborhood of $x$, then the function 
$$ g(y) = \sum_{|\alpha| < r} \frac{1}{\alpha!} f^{\alpha}(x)(y-x)^\alpha $$
obtained by taking the first $(r-1)$ terms in the Taylor series of $f$ will be $(r-1)$-tangent to $f$ at $x$, and will have vanishing $r$ derivatives in the coordinate system. The problem is that if you take a change of coordinates, with the exception where $f$ is the constant function, you can always find a different coordinate system in which the coordinate partial derivatives of $g$ of order $r$ is non-zero. 
In other words, "generically, the vanishing of all the $r$-th order derivatives is not coordinate invariant for $r > 1$".
A: Because you're working with manifolds and not vector spaces. If you change coordinates, for $k>1$, the $k$th derivative is only tensorial modulo lower derivatives. 
