Let $M$ be a compact manifold and $E$ a complex vector bundle. We will consider differential operators $P$ acting between $\Gamma^{\infty}(M,E)$. Let $\mathcal{P}$ be the algebra of all differential operators: then $\mathcal{P}$ is *filtered*. Once we have filtered algebra, we can associate the *graded* algebra $\mathcal{S}$ as follows: $\mathcal{S}^d:=\mathcal{P}^d/\mathcal{P}^{d-1}$. Note that $\mathcal{P}^0=\mathcal{S}^0=C^{\infty}(M)$. When $P$ and $Q$ have degrees $m$ and $n$ (resp.) then $[P,Q]$ has degree $m+n-1$. In particular $[P,f]$ has degree $m-1$ for a smooth function $f$. From this it follows that $C^{\infty}(M)$ is central in $\mathcal{S}$. This should allow us to realize elements in $\mathcal{S}$ as smooth sections of some vector bundle $U$: here $U\_x$ is defined as $\mathcal{S}/I\_x \mathcal{S}$ where $I\_x$ is the ideal of all smooth functions vanishing at $x$. 



> 
> Why $\mathcal{S}$ is isomorphic with $\Gamma^{\infty}(M,U)$?
> 
> 
> 



Elements of graded algebra associated with the algebra of differential operators as smooth sections
The vector space $U\_x$ will be infinite dimensional, so it's not immediately clear what $\Gamma^\infty(M,U)$ denotes. I assume you mean $\Gamma^\infty(M,U):=\bigoplus\_k \Gamma^\infty(M,U^k)$ where $U^k\_x:=\mathcal{S}^k/I\_x\mathcal{S}^k$. Then the question might be: Why is $\mathcal{S}^k$ isomorphic to $\Gamma^\infty(M,U^k)$? When $\mathcal{S}^k$ is projective an finitely generated over $C^\infty(M)$ (which I think it is in your setting) the answer is the [Serre-Swan theorem](https://en.wikipedia.org/wiki/Serre%E2%80%93Swan_theorem). You can find a proof for the $C^\infty$ setting in the book Nestruev, Smooth manifolds and observables.


**Added in response to the comment:** 


You wrote: 



> 
> Serre-Swan's theorem gives you that $\mathcal{S}^k$ can be identified with $\Gamma^\infty(M,V)$ for *some* vector bundle $V$ (provided we can show that $\mathcal{S}^k$ is finitely generated and projective). 
>  On the other hand, here we would like to have a concrete description of $V$ as $U$ where the fiber over $x$ is $\mathcal{S}^k/I\_x\mathcal{S}^k$.
> 
> 
> 


That concrete description *always* follows from an identification $\Gamma^\infty(M,V)=\mathcal{S}^k$. To be precise: from a vector bundle $V$, a $C^\infty(M)$-module $\mathcal{P}$ and a $C^\infty(M)$-module isomorphism $\phi: \Gamma^\infty(M,V)\to \mathcal{P}$, we always obtain a natural identification $V\_x= \mathcal{P}/I\_x\mathcal{P}$. This identification is constructed as follows: given $x\in M$, let's consider $\mathbb{R}$ as a $C^\infty(M)$-module via the canonical identification $C^\infty(M)/I\_x = \mathbb{R}$. Next we have a natural identification
$\mathcal{P}/I\_x\mathcal{P}=\mathcal{P}\otimes\_{C^\infty M} \mathbb{R}$, given in the direction 
$
\mathcal{P}/I\_x\mathcal{P}\to \mathcal{P}\otimes\_{C^\infty M} \mathbb{R}
$
by
$$
p \text{ mod } I\_x\mathcal{P}\mapsto p\otimes 1
$$
and in the other direction by
$$
p\otimes c \mapsto p\cdot c \text{ mod } I\_x\mathcal{P}.
$$
(I skip the details of verifying that these maps are well defined and yield the identity when composed. Feel free to ask if something is not clear.)


Hence, tensoring the isomorphism $\phi$ with $\mathbb{R}$ we obtain an identification 
$$
\alpha: \Gamma^\infty(M,V)/I\_x\Gamma^\infty(M,V) \to \mathcal{P} /I\_x\mathcal{P}.
$$
As last step, it's not hard to see that $\Gamma^\infty(M,V)/I\_x\Gamma^\infty(M,V)$ is naturally identified with $V\_x$
for any vector bundle $V$ (Corollary 11.9 in Nestruev).


To see that your concrete module $\mathcal{S}^k$ is finitely generated and projective we can construct an isomorphism $\mathcal{S}^k = \mathcal{Q}^\*\otimes\_{C^\infty(M)} \mathcal{Q}\otimes\_{C^\infty(M)} S^k D$, where I'm using the following abbreviations


* $\mathcal{Q} =\Gamma^\infty(M,E)$
* $D$ denotes the module of vector fields on $M$, i.e. sections of the tangent bundle,
* $S^k$ denotes the $k$-th symmetric product of a $C^\infty(M)$-module.
* $\mathcal{Q}^\*=\text{Hom}\_{C^\infty M}(Q,C^\infty M)$,


In one direction this iso is defined by
\begin{align}
 \mathcal{Q}^\*\otimes\_{C^\infty M} \mathcal{Q}\otimes\_{C^\infty M} S^k D &\to \mathcal{P}^k/\mathcal{P}^{k-1}\\
 r\otimes s \otimes (X\_1\cdots X\_k) &\mapsto s\circ X\_1\circ\ldots\circ X\_k \circ r \quad \text{mod} \quad \mathcal{P}^{k-1}
\end{align}


To make sense of the composition with $s$, use the canonical identification of a module $\mathcal{Q}$ with $\text{Hom}\_{C^\infty M}(C^\infty M, \mathcal{Q})$.


The other direction is also not to hard to define, but slightly more involved: start with the observation that given $(\nabla \text{ mod } \mathcal{P}^{k-1}) \in \mathcal{P}^k/\mathcal{P}^{k-1}$ and $a\_1,\ldots,a\_k \in C^\infty M$, then $[\cdots [\nabla,a\_1],a\_2]\cdots ,a\_k]\in \mathcal{Q}^\*\otimes \mathcal{Q}$ is well-defined (independent of the representative $\nabla\in \mathcal{P}^k$) and multilinear symmetric and a derivation in the $a\_i$'s. From there you can use the universal derivation $d: A \to \Lambda^1$ (differential one forms) to show that you get a well defined map from $ \mathcal{P}^k/\mathcal{P}^{k-1}$ to $\mathcal{Q}^\*\otimes\_{C^\infty(M)} \mathcal{Q}\otimes\_{C^\infty(M)} S^k D$.


This identification not only shows that $\mathcal{S}^k$ is projective and finitely generated, it gives a more concrete description of fibers $\mathcal{S}^k/I\_x\mathcal{S}^k$: they are symbols $S^k T\_xM\otimes\_\mathbb{R} \text{Hom}\_\mathbb{R}(E\_x,E\_x)$, as mentioned in Peter Michor's answer.


