Let $M$ be a smooth manifold of dimension $n$. Let $E$ denote the bundle of germs of smooth functions on $M$. For every stalk $E_x$ we can define the ideal
$$I_x^k=\{ f \in\mathcal{C}^{\infty}(M) \vert \text{ whose derivatives of order } \leq k \text{ vanish at } x \}$$
Then we can take the quotient $E_x/I_x^k=E_x^k$ which can be identified with the set of germs modulo the equivalence relation that identifies two germs iff their Taylor polynomials coincide in the first $k$ terms.
I want to show that the bundle $E/I$ is locally trivial. I don't really know how to do this. I guess we should pick a local chart $U$, and try to show that for every $x,y \in U$ we have $E_x^k \cong E_y^k$ but I am not sure.
I tried the trivial case $M=\mathbb{R}$. In this case I identify each fiber $E_x^k$ with the vector space of polynomials of degree $\leq k$ and I think I can show that $E^k$ is a trivial bundle. Is the general proof a generalization of this idea using local charts?
I assume you mean in the definition of $I^k_x$ that the derivative of order $k$ vanish at $x$.
You just need to analyze the local situation, for example, assume that $U$ is an open ball of $R^n$ containing $0$, then for every $x\in U$, you have the map $s_x:E_x\rightarrow E_0$ defined by $s_x(f)(y)=f(y+x)$ this map sends $I_x^k$ to $I_0^k$ and defines the trivialization $h:(E/I)_{\mid U}\rightarrow U\times E_0/I_0^k$ by $h(f)=(x,s_x(f))$ where $f$ is an element of the fibre of $x$.