If a language is not Turing reducible to two languages, may it still be Turing reducible to their "union"? Consider a language $L$ that is undecidable relative to $L_1$ and is also undecidable relative to $L_2$. Suppose, however, that there is a "multi"-oracle Turing machine $M$ that can query both the $L_1$ oracle as well as the $L_2$ oracle such that $M$ decides $L$. In other words, $L$ requires both the $L_1$ and $L_2$ oracles to be decided, but neither one alone suffices to decide $L$. I would like to come up with such languages $L$, $L_1$, and $L_2$. At first, I was hoping that the oracles for the rejecting and the accepting problems for Turing machines together could be queried to decide the halting problem (but neither oracle alone could be used decide it), but that turns out to be a dead end.

Now I'm wondering if I can approach the problem by trying to come up with two languages such that neither is Turing-reducible to the other, but I'm struggling to find examples.
A classical theorem of Sacks states that if $L$ is not computable, then it is almost surely not computable relative to a random oracle. In other words, if $O$ is a random oracle, then the probability that $L$ is computable given $O$ is zero.

Now take your favorite uncomputable $L$, and choose $L_1,L_2$ at random among all languages such that $L = L_1 \Delta L_2$ (here $\Delta$ is symmetric difference). Individually, $L_1,L_2$ are random oracles, and so almost surely $L$ is not computable given just $L_1$ or just $L_2$. However, it is clearly computable given both.

Another solution is to take any two incomparable Turing degrees $L_1,L_2$ and $L = \{ 0x : x \in L_1 \} \cup \{ 1x : x \in L_2 \}$. The proof then follows practically by definition.