You can get a slightly better upperbound than $\overrightarrow{\chi_{u}} (G)\leq  2^{\Delta(G)}$ by using sets of integers with distinct subset sums.  For example, [Bohman][1] constructed a set $S$ of $n$ positive integers with $2^n$ distinct subset sums and with maximum element less than $0.22002\cdot2^{n}$.  By taking a proper edge-colouring and assigning labels from $S$ rather than powers of $2$, we have that $\overrightarrow{\chi_{u}} (G)\leq 0.44004 \cdot 2^{\Delta(G)}$. 

You can also get slightly better lowerbounds as follows. Define a set $S$ of positive integers to be *good* if for all $s \in S$, there does not exist a set $T \subseteq S \setminus \{s\}$ such that $s = \sum_{t \in T} t$. For all $n \in \mathbb{N}$, let $$g(n)=\min \{\max S : \text{$S$ is a good set of $n$ positive integers}\}.$$

**Claim.** For all graphs $G$, 
$$\overrightarrow{\chi_{u}} (G) \geq  g({\Delta(G))}.$$

*Proof.* Let $x$ be a vertex of $G$ of maximum degree and let $S$ be the set of $\Delta(G)$ labels that appear on the edges incident to $x$. By definition, $S$ must be a good set.  Therefore, $\overrightarrow{\chi_{u}} (G) \geq \max S \geq g(\Delta(G))$, as claimed.  
 
I initially thought that $g(n)$ might be superpolynomial in $n$, but I now realize that $\{n-1, n, \dots, 2n-2\}$ is a good set of size $n$.  Therefore, $g(n) \leq 2n-2$.

 

  [1]: https://www.combinatorics.org/ojs/index.php/eljc/article/view/v5i1r3