OK, in fact this is easy: simply first prove that
$\sum_{j=k}^{2k}\binom{k}{j-k}\frac{B_{j+1}}{j+1}=(-1)^k\binom{2k}{k}\frac{1}{4k+2}$, the rest is immediate.

**Edit.** Henri, I'm editing. If you don't agree, please delete it.
We need to prove that
$$\sum_{j=k}^{2k}\binom{k}{j-k}\frac{B_{j+1}}{j+1}=\frac{(-1)^{k-1}}{\binom{2k}{k}(4k+2)}.$$
We also need to prove that
$$\sum_{k\geq0}\frac{(-1)^{k-1}}{\binom{2k}{k}(4k+2)}=\frac{2\log\phi}{1-2\phi}.$$