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:** I assume your $M$ is a real smooth manifold and your $C^\infty(M)$ denotes the algebra of real valued smooth functions. Further I assume your notation $\Gamma^\infty (M,V)$ means the $C^\infty(M)$-module of smooth sections of a finite dimensional vector bundle $\pi:V\to M$. 

Note that to define a vector bundle it's not enough to say what each fiber $V_x$ is. So in you're question you have not defined a vector bundle $U$.

Maybe the following facts can nevertheless help you. (These facts don't require $M$ to be compact. I suspect they only require minimal modification in case you consider complex valued smooth function. But since you were not very specific about the setting in your question, it is hard to provide a specific answer.):

 1. If $\pi: V\to M$ is a finite dimensional vector bundle, the natural $\mathbb{R}$-linear map of type $\Gamma^\infty (V)\to V_x$ which evaluates a section at $x\in M$ is surjective and has kernel $I_x\Gamma^\infty (V)$. So $\Gamma^\infty (V)/I_x\Gamma^\infty (V)=V_x$. (This is Lemma 11.8 in Nestruev)


 2. For $\pi: V\to M$ a finite dimensional vector bundle, $\Gamma^\infty (M,V)$ is finitely generated and projective as $C^\infty(M)$-module. Conversely, if $Q$ is a finitely generated projective $C^\infty(M)$-module, then there exists a finite dimensional real vector bundle $\pi:V\to M$, such that  $Q$ is isomorphic to $\Gamma^\infty(M,V)$, as $C^\infty(M)$-module. (Theorem 11.32 in Nestruev.)

These two facts answer the question in your last comment. It seems to me that centrality of $C^\infty(M)$ is a red herring here. (Also note that your claim $\mathcal{P}^0=C^\infty(M)$ is only true when $E$ is one dimensional. In general $\mathcal{P}^0=\Gamma^\infty(M,\text{End}_\mathbb{R}(M))$).

Finally, to show that $\mathcal{S}^k$ is finitely generated and projective you can construct an isomorphism $\mathcal{S}^k = \mathcal{Q}^*\otimes_A \mathcal{Q}\otimes_A S^k D$, where I'm using the following abbreviations


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

In one direction this iso is defined by
\begin{align}
 \mathcal{Q}^*\otimes_A \mathcal{Q}\otimes_A 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}

This yields, that the fiber $U^k_x$ of the bundle you are intersted in is 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.