I would write the following as a comment, but it was too long to fit in.
Let's first describe the meaning of “algorithms in class $\bf C$ with an oracle for a language A.” (The need for this was pointed out by Tsuyoshi Ito). We will use the same convention used by Ladner and Lynch. The convention is well-described by Bennett & Gill:
$\bf{LOGSPACE}^A$ can be defined in various ways, depending on how the query tape is handled. We follow the conventions of Ladner and Lynch [LL]: The query tape is not charged against the space bound, but to keep it from being used as a work tape, the query tape is one-way and write-only, and is erased automatically following each query. (Simon [Si] treats the query tape as one of the work tapes, a two-way read/write tape that is charged against the space bound. The Ladner-Lynch definition is less restrictive and perhaps more natural, since for a random oracle $A \in \bf{LOGSPACE}^A$ holds with probability 1 for [LL] but not for [Si])
[LL] R. E. LADNER AND N. A. LYNCH, Relativization of questions about log space computability, Math. Systems Theory, 10 (1976), pp. 19-32.
[Si] J. SIMON, On Some Central Problems in Computational Complexity, Tech. Rep TR 75-224, Dept. of Computer Science, Cornell University, Ithaca, NY, 1975.
The standard definition of complexity classes of oracle machines is as follows: Let B and C be complexity classes. Then, $X = B^C$ is a legitimate complexity class, defined as $X = \bigcup_{L\in C}{B^L}$. Here, $B^L$ represents the complexity class of decision problems solvable by an algorithm in class B with an oracle for a language L.
Since X is a legitimate complexity class, for any complexity class A, we can speak of complexity classes $A^X = A^{(B^C)}$ and $X^A = (B^C)^A$.
$A^X$ refers to the complexity class of decision problems solvable by an algorithm in class A with an oracle for any language $L' \in X = \bigcup_{L\in C}{B^L}$. In other words, $A ^ X = \bigcup\limits_{L'\in \{\bigcup\limits_{L\in C}{B^L}\}}{A^{L'}}$.
$X^A$ refers to the complexity class of decision problems solvable by an algorithm in class $X = \bigcup_{L\in C}{B^L}$ with an oracle for any language $L' \in A$. In other words, $X ^ A = \bigcup_{L' \in A} {{X^{L'}}} = \bigcup_{L' \in A} {{{\left( {\bigcup_{L \in C} {{B^L}} } \right)}^{L'}}}$.
Claim: $(B^{L_1})^{L'} \cup (B^{L_2})^{L'} = (B^{L'})^{L_1 \cup L_2}$.
Side Note: Since it's 3:00 AM now, I'm too sleepy to check the validity of the above claim! I think it's valid & elementary to prove, yet it's nice to see the actual proof.
Therefore, one can write $X ^ A = \bigcup_{L' \in A} {{{\left( {\bigcup_{L \in C} {{B^L}} } \right)}^{L'}}} = \bigcup_{L \in C, L' \in A} {{{\left( B^{L'} \right)}^L}}$.
Example
Let $\bf X = \bf{P^{NP}}$.
We know that $\bf{coNP} \subseteq \bf X$. Giving both sides oracle access to $\bf {NP}$, one gets: $\bf{coNP^{NP}} \subseteq \bf{X^{NP}} = (\bf{P^{NP}})^{\bf{NP}}$.
Epilogue
A fruitful discussion with Tsuyoshi Ito revealed (for me) that it is not easy to doubly relativize a complexity class. In fact, even defining one seems to be problematic. I should definitely study more to see if any satisfactory definition is ever given. Meanwhile, I appreciate any comment which can be used to solve this problem.
