The short answer is that you should write the state prior to measurement in the basis of the operator you intend to measure. Basically, the probabilities care about the coefficients of the state in the measurement basis only, so no other basis matters to computing the measurement probabilities. If you intend to measure a sequence of noncommuting operators, then consider the basis of the first measurement first, and then use the fact that, following that measurement, the state will be an eigenstate of the measured operator, and use this to compute subsequent probabilities. But there is no notion of "collapse in different bases'' because you always "collapse" into an eigenstate of the operator you measured, which has a fixed basis!
Let me know if I've missed some aspect of your question.
Since you mentioned the Bell state, let's consider that. I'm going to use the way I think about measurements, which is guaranteed to hold by the Stinespring Dilation Theorem, and has many nice features.
As you wrote, we have
$$ \left| {\rm Bell} \right\rangle = \frac{1}{\sqrt{2}} \left( \left| 00 \right\rangle + \left| 11 \right\rangle \right) = \frac{1}{\sqrt{2}} \left( \left| ++ \right\rangle + \left| -- \right\rangle \right) \, ,~~$$
where $Z \left|0 \right\rangle = \left| 0 \right\rangle$, $Z \left|1 \right\rangle = - \left| 1 \right\rangle$ are the $Z$-basis eigenstates, and $\left| \pm \right\rangle = ( \left|0 \right\rangle \pm \left| 1 \right\rangle )/\sqrt{2}$ are the $X$-basis eigenstates, $X \left| \pm \right\rangle = \pm \left| \pm \right\rangle$. The $Y$-basis Bell state I think is weird, but it's enough to think about two noncommuting bases!
Now, "collapse" is a weird concept, and not one I really think is worth thinking about. A measurement is just an entangling interaction between the system and the detector. Suppose two qubits are prepared in the Bell state above, and we send the first qubit to Alice ($A$) and the second to Bob ($B$).
Suppose Alice and Bob plan to measure a Pauli operator on their qubits. Let's also label the detectors Alice and Bob use for their local measurements as $a$ and $b$, respectively. Suppose that both apparati are initialized in the "default" state $\left| 0 \right\rangle$, and that the $Z$-basis state of the apparatus encodes the measurement outcome $n=0,1$, corresponding to the eigenvalue $(-1)^n$ of the measured Pauli.
Then, if Alice measures the Pauli operator $\sigma^\nu_A$, this can be represented by the channel
$$ \mathsf{V}^{\vphantom{\dagger}}_{A} = \frac{1}{2}\left( \mathbb{1} + \sigma^\nu_A \right) \otimes \widetilde{\mathbb{1}}^{\vphantom{\dagger}}_a + \frac{1}{2}\left( \mathbb{1} - \sigma^\nu_A \right) \otimes \widetilde{X}^{\vphantom{\dagger}}_a \, ,~~$$
where the tildes denote an operation on the detector. This is exactly how a fluorescent measurement of the $Z$-basis state of an atomic qubit works. If the atom is in the state 0, it won't fluoresce, no photons are detected, and the detector remains in the state 0, matching the state of the qubit. If the atom is in the state 1, it will fluoresce, a photon is detected, flipping the state of the detector to 1, matching the state of the qubit. Bob's measurements are captured by
$$ \mathsf{V}^{\vphantom{\dagger}}_{B} = \frac{1}{2}\left( \mathbb{1} + \sigma^\mu_B \right) \otimes \widetilde{\mathbb{1}}^{\vphantom{\dagger}}_b + \frac{1}{2}\left( \mathbb{1} - \sigma^\mu_B \right) \otimes \widetilde{X}^{\vphantom{\dagger}}_b \, ,~~$$
where the operators $\mathsf{V}$ are unitary!
Now the important part: Measurements act almost trivially if we represent the initial state in the basis of the operator that one intends to measure. If both Alice and Bob intend to measure $Z$, we write their shared Bell state in the $Z$ basis. In other words, we have the initial state
$$ \left| \Psi_0 \right\rangle = \frac{1}{\sqrt{2}} \left( \left| 0 \right\rangle^{\vphantom{\prime}}_A \otimes \left| 0 \right\rangle^{\vphantom{\prime}}_B + \left| 1 \right\rangle^{\vphantom{\prime}}_A \otimes \left| 1 \right\rangle^{\vphantom{\prime}}_B \right) \otimes \left| 0 \right\rangle^{\vphantom{\prime}}_a \otimes \left| 0 \right\rangle^{\vphantom{\prime}}_b \, ,~$$
and after Alice's measurement, we have
$$ \left| \Psi_1 \right\rangle = \mathsf{V}^{\vphantom{\dagger}}_A \, \left| \Psi_0 \right\rangle = \frac{1}{\sqrt{2}} \left( \left| 0 \right\rangle^{\vphantom{\prime}}_A \otimes \left| 0 \right\rangle^{\vphantom{\prime}}_B \otimes \left| 0 \right\rangle^{\vphantom{\prime}}_a + \left| 1 \right\rangle^{\vphantom{\prime}}_A \otimes \left| 1 \right\rangle^{\vphantom{\prime}}_B \otimes \left| 1 \right\rangle^{\vphantom{\prime}}_a \right) \otimes \left| 0 \right\rangle^{\vphantom{\prime}}_b \, ,~~$$
and following Bob's measurement, we have
$$ \left| \Psi_2 \right\rangle = \mathsf{V}^{\vphantom{\dagger}}_B \, \left| \Psi_1 \right\rangle = \frac{1}{\sqrt{2}} \left( \left| 0 \right\rangle^{\vphantom{\prime}}_A \otimes \left| 0 \right\rangle^{\vphantom{\prime}}_B \otimes \left| 0 \right\rangle^{\vphantom{\prime}}_a \otimes \left| 0 \right\rangle^{\vphantom{\prime}}_b + \left| 1 \right\rangle^{\vphantom{\prime}}_A \otimes \left| 1 \right\rangle^{\vphantom{\prime}}_B \otimes \left| 1 \right\rangle^{\vphantom{\prime}}_a \otimes \left| 0 \right\rangle^{\vphantom{\prime}}_b \right) \, ,~~$$
so that
$$\left| \Psi_2 \right\rangle = \mathsf{V}^{\vphantom{\dagger}}_B \, \mathsf{V}^{\vphantom{\dagger}}_A \, \left| \Psi_0 \right\rangle = \frac{1}{\sqrt{2}} \left( \left| 0000 \right\rangle + \left| 1111 \right\rangle \right)^{\vphantom{\prime}}_{ABab} \, ,~~$$
so that nothing really happens to the physical part of the state, but the measurement guarantees that the device's state agrees with the state of the physical qubits in the basis measured!
If Alice measures $Z$ while Bob measures $X$, we should write the initial state in the $Z_A\otimes X_B$ basis,
$$ \left| \Psi_0 \right\rangle = \frac{1}{2} \left( \left| 0+ \right\rangle + \left| 0 - \right\rangle + \left|1 + \right\rangle - \left| 1 - \right\rangle \right) \, , ~~$$
so that, on measurement, we find that
$$ \left| \Psi_2 \right\rangle = \mathsf{V}^{\vphantom{\dagger}}_B \, \mathsf{V}^{\vphantom{\dagger}}_A \, \left| \Psi_0 \right\rangle = \frac{1}{2} \left( \left| 0+,0+ \right\rangle + \left| 0 -,0- \right\rangle + \left|1 +,1+ \right\rangle - \left| 1 -,1- \right\rangle \right)^{\vphantom{\prime}}_{ABab} \, ,~~$$
which, as before, is the same physical state with the outcomes (the detector's state) entangled to agree.
Now, where does collapse come in? Simply project any of the post-measurement states $\left| \Psi_2 \right\rangle$ above onto a particular outcome by projecting onto the state of the detector. In the case where both measure $Z$ (or $X$), notice that Alice and Bob's outcomes have to agree. They're either both 0 or both 1, with equal probability of 1/2. The probability is the expectation value of the detector state being 0, 1 (which is the usual Born rule, but it's moved to the detector now). Once you project onto an outcome, the resulting state is the "collapsed" state. In the case of $Z$ and $Z$ measurements, it's either $\left| 00 \right\rangle$ or $\left| 1 1 \right\rangle$. In the case of $Z$ and $X$ measurements, the two outcomes are independent, so there are four possibilities. If we do not project onto an outcome, you basically have the many-worlds wavefunction. Note that the measurement unitary is precisely the time evolution that one applies to the system + detector during the measurement process (there is only time evolution if you only consider closed systems, and life is easy).
But anyway, it's all quite transparent if you write the pre-measurement state in the basis you intend to measure. Everything I said above can be done with Kraus operators instead, though I find the formalism above more revealing.
