Consider a representation $a_1, a_2, \ldots, a_n$ of $t$ with the minimum number of non-zero terms. Suppose for the sake of contradiction that there are $l \geq 2k$ non-zero terms and enumerate them as $a_{i_1}, a_{i_2}, \ldots, a_{i_l}$. The main claim is the following: Claim: There exist disjoint non-empty subsets $S, T$ of $\{i_1, i_2, \ldots, i_l\}$ such that $\sum_{s \in S} u_s = \sum_{t \in T} u_t$. Proof: Let $I = \{i_1, i_2, \ldots, i_l\}$. Consider the map $f : \mathcal{P}(I) \to \mathbb{Z}$ defined by $f(X) = \sum_{x \in X} u_x$ for $X \subset I$. Since each $u_i$ is at most $2^k$, the image of $f$ is a subset of $\{0, 1, \ldots, l \cdot 2^k\}$, so it has size at most $l \cdot 2^k + 1$. We have $2^{\frac{l}{2}} > l$ (easily follows by induction on $l \geq 6$), so since $l \geq 2k$, it follows that $2^l > l \cdot 2^k$, and hence $2^l > l \cdot 2^k + 1$ due to parity. Thus, $|\mathcal{P}(I)| > |f(\mathcal{P}(I))|$, so $f$ isn't injective. Therefore, there exist distinct $X, Y \subset I$ such that $f(X) = f(Y)$. Then taking $S = X \setminus Y, T = Y \setminus X$ does the job. $\square$ Now take $S$ and $T$ as in the claim and let $m = \min_{s \in S} a_s$. By decreasing all $a_s$ for $s \in S$ by $m$ and increasing all $a_t$ for $t \in T$ by $m$ we obtain a representation with a smaller number of non-zero terms, which yields the desired contradiction.