How are frames of references compatible with quantum mechanics? The idea of changing the frame of reference is that you want to switch to the point of view of an object that is located at a well-defined position and moving with a well-defined velocity relative to you. However, quantum mechanics says that no object, including humans observers, can have a well-defined velocity and a well-defined position. For instance, we can't talk about the frame of reference of an electron, can we?
So how come a frame of reference is compatible with quantum theory? Do we modify the definition of a frame of reference from classical physics?
 A: Strictly speaking, individual objects don't have any such thing as position or velocity. A pair of objects has a relative displacement vector and a relative velocity vector. A large number of objects have a web of displacement vectors and relative velocity vectors.
It is mathematically convenient to assign one object to have zero displacement and velocity. Then we can express all the other displacement vectors and velocity vectors in terms of that object, the origin. (Or invent an imaginary object and assign it zero displacement and zero velocity, if that makes for easier math.) Then we express "position of the object relative to the origin" and "velocity of the object relative to the origin" for each of many objects. The practice is so common that the "relative to the origin" part is understood and we leave it out.
This can lead to a misunderstanding that velocity and position are properties of objects themselves. However, velocity and position are properties of a system of two objects and any uncertainty in those quantities therefore belongs not to one or the other object, but to the vector between them.
If you express the uncertainty of velocity and position as things belonging to the vectors between two objects, not to the objects themselves, I think you will find that your philosophical conundrum disappears.
A: 
"However, quantum mechanics says that no object, including humans observers, can have a well-defined velocity and a well-defined position."

This is the case when the objects are  treated as particles. A wavefunction can have a definite value (amplitude and phase) at every point, and a reference frame can be defined to align with conceptual, not-directly-measurable properties of the wave packet like the expectation (mean) position and its motion. The wave packet is broad and spreads out in a range of directions, so if you make a measurement of its position or velocity, you can get one of a range of values. But so long as you don't measure it, the wavepacket shape itself is not fuzzy. It's expectation position at each time is not fuzzy. And that expection position moves unfuzzily over time.
It's like any random variable. Say we have a jar of straws with lengths having a normal distribution. If we pick a straw, it could have any value in the spread. If we pick several hundred straws and average their lengths, that still only approximately estimates the true centre of the distribution. But the true centre of the distribution is a precise number. The expectation of a random variable is not itself a random variable, but a real number.
This can only be done conceptually, of course. Practically, we are always limited by measurement error, and so in practice we can only estimate or approximate the  reference frame following a real-world particle wavepacket. There will always be some residual error. But then that's true in the classical world too! No practical measurement of a continuous quantity is ever without error.
A: I'm not going to address relativistic quantum mechanics here as I don't really understand it, but I think this answer will cover the intuition for what you need in the classical space.
TLDR: Quantum Frames of Reference are probability distributions over position and momentum just like classical frames of reference are fixed coordinates of position and momentum. Changing frames of reference in quantum mechanics amounts to convolution as opposed to merely subtraction of vectors in classical mechanics.
See below: (I have edited this all to 1 dimension and only position to make it easier to read, please leave a comment if you care for elaboration on momentum + multiple dimensions)
Two objects in classical mechanics
Entities: the observer, some object $O_1$ some object $O_2$.
In classical mechanics we describe the state of the objects $O_1$ and $O_2$ using explicit vectors which describe where they are located and how they are moving relative to our observer. Since we are working in 1 spatial dimension our vectors will consist of a single number inside parenthesis.
So we can say $O_1$ has location $(x_1)$ relative to the observer. We can repeat this for the object $O_2$ by saying that $O_2$ has location $(x_2)$ relative to the observer.
We might then decide to ask "What is $O_2$'s state relative to $O_1$?". At which point we use our intuition of "frames of reference" to take $O_2$'s location and "subtract away" $O_1$'s frame of reference yielding:
$ (x_2 - x_1) $ as the new position (this is just $O_2$'s coordinates - $O_1$'s coordinates).
Two Objects in Classical Quantum Mechanics:
Entities: the observer, some object $O_1$ some object $O_2$
In quantum mechanics we cannot assign explicit states to $O_1$ and $O_2$ the best we can do is assign wave functions. We will define the following
$$ \begin{matrix} \Psi_1(x) : \text{position wave function of object 1} \end{matrix} \\ \begin{matrix} \Psi_2(x) : \text{position wave function of object 2} \end{matrix}$$
Reading the state of an object off of its wave function is a little tricky. We can consider a SET of positions $D$ and we have the following calculation we can do (via integration):
$$ \int_{D} |\Psi_1(x)|^2 = \text{Probability of finding the location of object 1 in D} $$
$$ \int_{D} |\Psi_2(x)|^2 = \text{Probability of finding the location of object 2 in D} $$
So now just like the classical case, we might be interested in asking "What is the relative probability distribution of $O_2$ compared to $O_1$"? The answer to this is much tougher to find. You can't just subtract the wave functions as we had subtracted frames of reference in the classical case.
You need to break the problem down by as follows:
Suppose we "know" the position of $O_1$ call it $(x_1)$ (even though this is unrealistic), the relative position probability distribution of $O_2$ would still be identical in shape, but it will be shifted in coordinates around this point. If we denote $\Psi_{2, \text{relative}}$ as the shifted probability distribution we then would have:
$$|\Psi_{2, \text{relative}}(x)|^2 = |\Psi_{2}(x - x_1)|^2 $$
Look at this carefully and convince yourself that the shape of the distribution HAS NOT changed, only the location. This is what we want from a "change of reference" ideally.
Now we aren't done, what we want is to generalize this principle to $O_1$ itself having a completely probabilistic location (it is a quantum object after all, whose TRUE state is described using $|\Psi_1|^2$).  We know the probability density of a particular location $(x_1)$ is given by $|\Psi_1(x_1)|^2$ and we know that at this location our $O_2$ probability distribution would look like $|\Psi(x - x_1)|^2$, and we want to sum over ALL the locations weighted by probability density, so its natural to choose to integrate and that brings us to this answer:
$$|\Psi_{2, \text{relative}}(x)|^2 = \int_{x_1}  |\Psi_2(x - x_1)|^2 |\Psi_1(x_1)|^2 dx_1  $$
This is also called a "convolution" which you may have learned in fourier analysis or in probability theory.
Similar such calculations can be carried out for the momentum wave functions.
So now heres the full picture:
In the classical world: Objects have definite position and momentum vectors, and calculating relative state comes down subtracting "frames of reference" which means subtracting position and momentum vectors.
In the quantum world: Objects have wave functions of position and momentum vectors. Calculating relative probability distributions comes down to convoluting pairs of wave functions.
So to finally answer your question: "Quantum Frames of Reference" are best understood as wave functions themselves and the underlying transformation changes from subtracting coordinates to convolution.
Additional things for consideration:

*

*Subtracting coordinates is EQUIVALENT to taking convolutions with delta functions. See if you can prove that the classical case is equivalent to the quantum case but replacing all probability distributions with delta functions.


*Note that in the quantum world we CANNOT calculate the RELATIVE WAVE FUNCTION itself (with complex numbers), ALL we can calculate is the relative PROBABILITY DISTRIBUTION, if you look above, EVERY SINGLE $\Psi$ term came with an absolute value squared around it. I never had a naked $\Psi$ running around. That is very intentional and needs to be digested.
A: A reference frame does not have to be attached to a physical object - it can be simply an abstract system of co-ordinates.
Provided we restrict ourselves to reference frames that have constant velocities relative to one another (i.e. the realm of special relativity) then it is possible to create a relativistic version of the Schrödinger equation This was discovered by Paul Dirac and is called the Dirac equation. It is also possible to create a version of quantum field theory in curved spacetime in which matter fields are quantised although the gravitational field is still classical - this is known as semi-classical gravity.
However, we do not yet know how to formulate a theory of quantum gravity, a combination of quantum mechanics and general relativity that is completely compatible with both theories.
