Interpretation of quantum superposition and classical Brownian motion In the standard, Copenhagen interpretation of quantum mechanics, the usual ontology assigned to the phenomenon that repeated measurements of a quantum mechanical observable yielding different results with various probabilities is that the system, due to its being in a quantum superposition before measurement, must "exist in many classical states at once". 
To be concrete, suppose the system I have in mind is a single particle (atom/ion/molecule) propagating in a double-well potential and the observable I'm measuring is the position of the particle, but that I really only care about which well (right or left) the particle shows up in upon measurement. Before measurement, the standard interpretation tells me that the particle is literally occupying both wells at once.
Now suppose I take the classical thermal analog of the above system. That is, I confine a classical particle to propagate in a double-well potential and immerse it in a thermal bath so that its trajectory is stochastic/Brownian due to the random bombardment by molecules/degrees of freedom comprising the bath. If I record the position of such a classical particle at times separated by intervals much larger than the oscillation period of the wells, then, due to the meta-stability of potential energy maximum between the wells, I will either find the particle in the left well or the right well with various probabilities (these probabilities will be equal for the symmetric double-well, as will also hold for the quantum case above). 
In this classical thermal situation, we would never say that the particle is occupying both wells at once. This is due to the fact that we can use light waves to observe the particle continuously without affecting its dynamical state. In the quantum setting, however, any attempt to observe the system causes an abrupt disturbance to the dynamical state. I like to call this issue in the quantum case "intrusive measurement".
My question has two parts:


*

*why must we say that a quantum particle is in two places/states at once just because we observe different measurement outcomes with various probabilities? 

*Considering the classical thermal analog above as a starting point, can we not say that there exist fluctuating degrees of freedom that cause quantum systems to exhibit random outcomes upon measurement?
 A: Because it doesn't agree with experiment
A particle in a superposition of states isn't in one well sometimes and in the other well at other times, randomly switching back and forth. QM says that the particle is in both wells simulatneously.  For example, if a charged particle randomly flops back and forth in a double well system that is symmetric about the origin, it will have a fluctuating dipole moment measured w.r.t. the origin.  But an electron in the ground state of the potential has zero dipole moment because it's not on one side or the other.
A: 

*

*why must we say that a quantum particle is in two places/states at once just because we observe different measurement outcomes with
various probabilities?


Physics does not say that, this is just a pop culture representation. A particle in a superposition of locations is not in both locations at once.
The proper way to think about a superposition is that it does not translate at all in the classical worldview. We just don't know where a superposed particle is, or even if it is a meaningful question to ask, because quantum mechanics is not counterfactual: there is no outcome to a measurement that is not performed. And when we do perform a measurement we never get anything suggesting that a particle may be at two locations at once, unless we look at the probability distribution of outcomes where interference patterns may appear, which leads me to the second point...



*Considering the classical thermal analog above as a starting point, can we not say that there exist fluctuating degrees of freedom
that cause quantum systems to exhibit random outcomes upon
measurement?


No, because this would not explain interference effects. A fluctuating value is well-defined at any time, so we have no reason to model it with probability amplitudes - a probability density would suffice. This would not produce interference patterns.
A: The way you can, and must, distinguish between a superposition and a classical probability distribution is that you must do something that causes quantum interference between its terms or, to say another way, you must ultimately do a measurement in a vector basis that is incompatible (the operator it allies with does not commute with) to the one you are interested in.
The mathematical formalism of quantum mechanics can be considered in many ways as a generalization of probability theory: such a concept is called in foundational papers a generalized probability theory or GPT, and we can thus make a point-by-point comparison with classical probability theory. In particular, the analogy of an ordinary probability density function is the quantum state, and the analogy of the Bayesian probability update is "measurement". Note that if you take the GPT view, this pretty much fixes the interpretational issues: states are indeed knowledge/information states and "wave function collapse" is indeed proximately nothing more than a subjective information acquisition. The two terms occupy mathematically exactly the same points in their respective theories.
What makes things "interesting" is their behavior, because after all, it is a generalized probability theory. You see, when you have a classical probability density function or pdf, like say the usual "bell curve"
$$f_X(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$$
it is implicitly understood that it only provides information, or models an agent's held information rather, regarding the variable $X$, and any functions of $X$ alone. On the other hand, a wave function, which by the way is only the extreme case of quantum states (and in non-Hilbert formalisms, the distinction doesn't exist explicitly, but it's still useful for analysis), does not just carry information about a single variable $X$, but also carries information about at least one more variable, e.g. in the case of a moving free particle, we also have its momentum, $\mathbf{p}$, determined by the same wave function but in a different way than, say, a mere function of the classical $X$ above would, such that the wave functions that localize $X$ to a point are mutually exclusive with those that localize $\mathbf{p}$ to a point, so no information gain can ever reduce your wave function state to one where both are localized on a point. In a sense, squeezing information into one squeezes it out of the other, and conversely. In particular, if we instead have
$$\psi_x(x) = \frac{1}{(2\pi)^{1/4}} e^{-x^2/4}$$
and we receive information that $x = x_0$, we must change to
$$\psi'_x(x) = \delta^3(x - x_0)$$
but if we receive that $p = p_0$, we must change to
$$\psi'_x(x) = e^{ip_0 x/\hbar}.$$
But note! If we were not allowed to measure $p$, or better, $p$ "didn't exist", and our scenario only included $x$, then in fact the behavior of $\psi_x$ is identical to classical probability theory, and since the proper "probability" is it norm-squared, we can apply this theory to every single instance we can apply classical probability to without any issue at all! It just becomes oddly redundant/complicated, with the additional complex phase term (not invoked in the above).
Thus we can return to your question. For a classical Brownian motion system of two wells, you are right, we can set up a suitable setup in which an agent using classical probability theory is justified in describing its knowledge via
$$f_x(x) = \frac{\mathrm{rect}\left(x - 2\right) + \mathrm{rect}\left(x + 2\right)}{2}$$
where $\mathrm{rect}$ is the "rectangle" or "unit pulse" function, a uniform discontinuous step one unit wide and high, zero everywhere else, and we imagine it as lying in a pair of potential wells of suitable width sunk around $x$-coordinates $\pm 2$. With it energetic enough, it flits over the barrier and produces probability in both. There is also mathematically NO difference between this and the wave function
$$\psi_x(x) = \frac{\mathrm{rect}\left(x - 2\right) + \mathrm{rect}\left(x + 2\right)}{\sqrt{2}}$$
so long as we are only measuring $x$ and functions of $x$ alone!
The problem is, in a real quantum setup, we can also measure $p$.
And then, there's a difference. In fact, it gets even weirder - we can suitably construct a hybrid observable, $q$, that is a function of both $x$ and $p$ such that if $q$ takes a certain value, the "update rule" must set your $\psi_x$ to the above function. In particular, when it already is the above function, that observable must yield that answer with probability 100%!
Now think about that in light of the classical Brownian system. What that is saying is equivalent to saying that although the Brownian particle is only in one well or another at any given time, we had a measurement we could run that would be equivalent to affirming that its probability distribution for all time, given  only its state at that single moment in time, 100% for sure, is some certain way! How would you do that? If you just know right now it is dead center of the left well, say, how can you measure just from that information alone that its probability density is the above shape and not some other shape? Remember that probability density only appears in aggregate repetition, not just one experiment, yet in quantum theory, we have at least the potential to verify it with one experiment alone!
Of course, if you're astute enough, you might protest "but we can know the future probability! If I know $x$ and the velocity of the ball I can predict all places it goes and thus reconstruct from present information that yes, '100%' it has that distribution for any future sampling". Alas, though, that doesn't work: with a suitably clever setup, such as that used in Bell's Theorem, you can find that while such a thing makes sense conceptually, the precise statistics it comes up for for all such processes in aggregate just don't match the quantum ones, no matter how cleverly you imagine whatever is going on "behind the scenes". And when we do the experiment on real life systems, the quantum statistics win out. Every last time.
Instead, it seems as if $\psi_x$ somehow represents, itself the actual information that is present in reality, as well as your knowledge thereof, and not just one or the other alone, under the hypothesis of the existence of an incompatible variable.
