The definition of an oracular complexity class is key to answering this question, it is often defined as:
$$
\mathsf{C^{A}=\bigcup_{L_A\in A}C^{L_A}}
$$
But how do we define two oracles? Well the union definition ($\mathsf{C^{A,B}}=\mathsf{C^{A\cup B}}$) would imply that a Turing machine has access to only one language from either $\mathsf{A}$ or $\mathsf{B}$, which is not quite what we want. We would like that it has a language from both:
$$
\mathsf{A^{B,C} = \bigcup_{L_B\in B,\hspace{0.05cm}L_C\in C} A^{L_B, L_C}}.
$$
Choosing this definition would imply that $\mathsf{A^{B,C}\not = \mathsf{C^A \cup C^B}}$ either. For counter-example assume there exists some $\mathsf{L_A\in C^A\setminus C^B}$ and some $\mathsf{L_B\in C^B\setminus C^A}$. Deciding the language:
$$
\mathsf{L}=\{x\#y: x\in L_A, y\in L_B
\}
$$
is in $\mathsf{C^{A,B}}$ under our definition (assuming C is capable of Turing reductions), but it is in neither $\mathsf{C^A}$ or $\mathsf{C^B}$, so it can not be in $\mathsf{C^A \cup C^B}$.
So what is $\mathsf{C^{A,B}}$? Well, it seems to be it is its own object, without a simple equality to some other understood concept. Although, it often can be simplified when $\mathsf{A}$, $\mathsf{B}$ or $\mathsf{C}$ are known. For example, in the case you state there is a simple reduction.
$$
\mathsf{P^{NP,CoNP}=P^{NP}
}.
$$
This is because $\mathsf{P^{NP}=P^{CoNP}}$ and polynomial time TMs can simulate other polynomial time TMs.