To elaborate on the above comment, the problem with order $k$ bases is precisely that Chernoff's inequality does not work. The joint independence assumption for Chernoff's inequality is essential; as seen by the following example taken from Tao and Vus' Additive Combinatorics:
Color the elements of $[1, N]$ either black or white independently and with equal probability. For each $A \subset [1, N]$ let $s_A$ denote the parity of black elements of $A$ (so say if $A$ contains 3 black elements then $s_A = 1$). One can check that the $s_A$'s are independent events. Write $X = \displaystyle \sum_{A \subset [1, N]} s_A$. One can check that $\mathbb{E}X = 2^N - 1/2$ and $\textbf{Var} X = 2^{N-2} - 1/4$. Further, $\mathbb{P}(X = 0) = 2^{-N}$. The upper-bound on Chernoff's inequality would be $2\exp(-2^{N-2})$, which is much smaller than $\mathbb{P}(X = 0)$, so the inequality fails.
The reason why a simple argument suffices for additive bases of order 2 is because we have
$$\displaystyle r_{2,B}(n) = \sum_{x < n/2} \mathbb{I}(x \in B) \mathbb{I}(n - x \in B) + E$$
where $E$ is a suitably small error, and $r_{2,B}(n)$ is the number of ways to write $n$ as a sum of two elements in $B$. The key here is that the events $\mathbb{I}(x \in B) \mathbb{I}(n - x \in B)$ are independent for $1 \leq x < n/2$. This is not the case when there are more summands. In the Erdos-Tetali paper cited above, this issue is circumvented via Janson's inequality. In particular, Erdos-Tetali showed that there are additive bases of order $k$ satisfying $| B \cap [1,N]| = \Theta(N^{1/k} \log^{1/k} N)$.
The main difficulty you have to circumvent is how to deal with the non-independence of the random variables $\mathbb{I}(x_1 \in B) \cdots \mathbb{I}(x_k \in B)$.