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 reduces to: 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). I think you can find a proof for the $C^\infty$ setting in the book Nestruev, Smooth manifolds and observables.

**Added after the comments:** Since you were not very specific about the setting in your question, I'm going to make the following assumptions: your $M$ is a real smooth manifold and $C^\infty(M)$ denotes the algebra of real valued smooth functions. I assume that $\Gamma^\infty (M,V)$ denotes the $C^\infty(M)$-module of smooth sections of a finite dimensional real vector bundle $\pi:V\to M$. You mention that the bundle $E$ is complex, but I'll assume, when you talk of differential operators between $\Gamma^\infty(M,E)$, you mean differential operators in the real sense, i.e. $P:\Gamma^\infty(M,E)\to \Gamma^\infty(M,E)$ is a differential operator of order $k$, iff $[f,P]$ is a differential operator of order $k-1$ for every $f\in C^\infty(M)$. I believe that what I'm about to say has straightforward modifications in case you meant $C^\infty(M)$ to be complex valued functions, or $M$ to be complex analytic.

In your 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 $V$ ...
On the other hand, here we would like to have a concrete description of $V$ as $U$ where the fiber over $x$ is $\mathcal{S}^k/I_x\mathcal{S}^k$.

The proof of Serre-Swan given in for example Nestruev (Theorem 11.32) indeed seems to only give *some* vector bundle, without showing that the construction is functorial. It's probably possible to show that the construction in Nestruev *is* functorial, but one could also address the concern in your comment directly, by defining the functor from projective finitely generated $C^\infty(M)$-modules to vector bundles over $M$ as
$$\mathcal{Q} \mapsto \left(\bigcup_{x\in M} \mathcal{Q}/I_x   \mathcal{Q}\to M\right).$$ 
To make sense of this, we still need to explain what topology and smooth structure to put on the disjoint union $\bigcup_{x\in M} \mathcal{Q}/I_x   \mathcal{Q}$. It should not be to hard to fill in the details with the other things you can find in the Nestruevs book. (Note that by saying what each fiber $U_x$ is in you're question, you had *not* defined a vector bundle $U$ yet.) 

Finally, to show that $\mathcal{S}^k$ is finitely generated and projective you 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

 - $Q =\Gamma^\infty(M,E)$
 - $D$ 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.
 - $Q^*=\text{Hom}_{C^\infty(M)}(Q,C^\infty(M))$.

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 P^k/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 P^{k-1}
\end{align}

The other direction of the iso is also not to hard to define, but is slightly more involved.

With this isomorphism we could have skipped all the talk of Serre-Swan, since it directly yields a description of the fiber $U^k_x$ of the bundle you are interested in. Namely it's canonically isomorphic to $S^k T_xM\otimes_\mathbb{R} \text{Hom}_\mathbb{R}(E_x,E_x)$, which are symbols, as Peter Michor mentions. (This final step requires the fact that pulling back modules (extension of scalars) commutes with tensor products.)