Is there something behind non-commuting observables? Consider a quantum system described by the Hilbert space $\mathcal{H}$ and consider $A,B\in \mathcal{L}(\mathcal{H},\mathcal{H})$ to be observables. If those observables do not commute there's no simultaneous basis of eigenvectors of each of them. In that case, in general if $|\varphi\rangle$ is eigenvector of $A$ it will not be of $B$.
This leads to the problem of not having a definite value of some quantity in some states.
Now, this is just a mathematical model. It works because it agrees with observations. But it makes me wonder about something. Concerning the Physical quantities associated to $A$ and $B$ (if an example helps consider $A$ to be the position and $B$ the momentum) what is really behind non-commutativity?
Do we have any idea whatsoever about why two observables do not commute? Is there any idea about any underlying reason for that?
Again I know one might say "we don't care about that because the theory agrees with the observation", but I can't really believe there's no underlying reason for some physical quantities be compatible while others are not.
I believe this comes down to the fact that a measurement of a quantity affects the system in some way that interferes with the other quantity, but I don't know how to elaborate on this.
EDIT: I think it's useful to emphasize that I'm not saying that "I can't accept that there exist observables which don't commute". This would enter the rather lengthy discussion about whether nature is deterministic or not, which is not what I'm trying to get here.
My point is: suppose $A_1,A_2,B_1,B_2$ are observables and suppose that $A_1$ and $B_1$ commute while $A_2$ and $B_2$ do not commute. My whole question is: do we know today why the physical quantities $A_1$ and $B_1$ are compatible (can be simultaneously known) and why the quantities $A_2$ and $B_2$ are not?
In other words: accepting that there are incompatible observables, and given a pair of incompatible observables do we know currently, or at least have a guess about why those physical quantities are incompatible?
 A: Observables don't commute if they can't be simultaneously diagonalized, i.e. if they don't share an eigenvector basis. If you look at this condition the right way, the resulting uncertainty principle becomes very intuitive.
As an example, consider the two-dimensional Hilbert space describing the polarization of a photon moving along the $z$ axis. Its polarization is a vector in the $xy$ plane.
Let $A$ be the operator that determines whether a photon is polarized along the $x$ axis or the $y$ axis, assigning a value of 0 to the former option and 1 to the latter. You can measure $A$ using a simple polarizing filter, and its matrix elements are
$$A = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.$$
Now let $B$ be the operator that determines whether a photon is $+$ polarized (i.e. polarized southwest/northeast) or $-$ polarized (polarized southeast/northwest), assigning them values 0 and 1, respectively. Then
$$B = \begin{pmatrix} 1/2 & -1/2 \\ -1/2 & 1/2 \end{pmatrix}.$$
The operators $A$ and $B$ don't commute, so they can't be simultaneously diagonalized and thus obey an uncertainty principle. And you can immediately see why from geometry: $A$ and $B$ are picking out different sets of directions. If you had a definite value of $A$, you have to be either $x$ or $y$ polarized. If you had a definite value of $B$, you'd have to be $+$ or $-$ polarized. It's impossible to be both at once.
Or, if you rephrase things in terms of compass directions, the questions "are you going north or east" and "are you going northeast or southeast" do not have simultaneously well-defined answers. This doesn't mean compasses are incorrect, or incomplete, or that observing a compass 'interferes with orientation'. They're just different directions. 
Position and momentum are exactly the same way. A position eigenstate is sharply localized, while a momentum eigenstate has infinite spatial extent. Thinking of the Hilbert space as a vector space, they're simply picking out different directions; no vector is an eigenvector of both at once.
A: You can make a connection to the fact that the exact physical state of an arbitrary physical system that occupies a finite volume, can be specified with only a finite amount of information. If you consider some observable then the eigenstates may be degenerate, you then need another observable commuting with the first one to lift that degeneracy, if you continue this way you'll eventually end up with a complete set of commuting observables. Since only a finite amount of information is need to specify the state of the system, this means that this set will be finite. It is then guaranteed that you can find observables that don't belong to this set.
In the classical limit, all observables commute. In this limit, the number of distinguishable physical states per unit phase phase tends to infinity. 
A: Classical mechanics can be expressed in analytical form in terms of position and "conjugate momentum", a term from Lagrangian mechanics. This pair gives us the P, Q variables of Hamiltonian mechanics; when any such pair is quantized you will find that they do not commute. The quantum commutator "inherits" this behavior from the classical Poisson brackets.
So this provides some physical background to your question; the Lagrangian ultimately derives from Newton's Laws of Motion via the Principle of Least Action, a variational principle.
A: Yes, there is a fundamental reason why some observables do not commute. Some are the non-commuting generators of a group. For example, the angular momenta $J_x,J_y,J_z$ are observables.  They are also the generators of rotations in the rotation group. By rotating a pencil with your fingers, you can verify that $Rot_xRot_y$ and $Rot_yRot_x$ do not yield the same pencil orientation.  Rotations do not commute and by considering small rotations you deduce the commutation relations $[J_k,J_l]=i\epsilon_{klm}J_m$.
A: You say that

This leads to the problem of not having a definite value of some
  quantity in some states.

but what sort of problem is this? It's not a problem of disagreement between theory and experiment. In fact, as I will get to, it's the other way around! If observables commute, we cannot construct theories that agree with experiment. So the problem here is that this is a psychologically or philosophically difficult thing to accept, but that's the way it is and if you don't like it you have to find another universe where the rules are simpler...
Bell's inequalities and GHZ experiments can be argued to show that there really isn't anything behind non-commuting observables.  A theory built on only commuting observables simply can't give the correct prediction for the GHZ experiment, but QM does (very easily, too). So non-commuting observables seem to be a very fundamental part of the way our universe works.
Whatever you would suggest is "really" behind quantum mechanics has to make the same prediction as quantum mechanics about the results of a Bell or GHZ experiment, because we have done those experiments and found the result predicted by QM. This rules out that if there is something behind quantum mechanics, all its observables commute.
The universe looks quantum mechanical because it is quantum mechanical.
(There's a loophole in the above: we could allow for faster-than-light signals and then there could be hidden variables. But that's genuinely disturbing to physicists in a sense that non-commuting observables isn't, because if we allow faster-than-light communication we must, according to Einstein, allow conditions in the future to affect what happens in the present. That means I can't trust my experiments anymore because someone from the future might be interfering with them. Non-commuting observables means we have to accept only predicting probabilities in experiments but that's much better than throwing all experiments out because your rival from the future could be sabotaging them.)
A: I'd like to expand a bit on a subtly incorrect statement at the beginning of the original post that was pointed out in the comments:

OP: If those observables do not commute there's no simultaneous basis of eigenvectors of each of them. In that case, in general if $|\phi \rangle$ is eigenvector of $A$ it will not be of $B$. (italics $\rightarrow$ incorrect statement)
Comment (WillO): You say that if observables do not commute, they can have no eigenvectors in common. This is not true.

To give the simplest concrete example, suppose our Hilbert space is finite-dimensional and $A$ and $B$ are non-commuting observables (matrices). Consider these two "expanded" operators on a similarly "expanded" Hilbert space:
$$A'=\begin{pmatrix}
A & 0 \\
0 & 1\end{pmatrix},\,\,B'=\begin{pmatrix}
B & 0 \\
0 & 1\end{pmatrix}$$ 
Clearly $A'$ and $B'$ also do not commute, but they do share an eigenvector $v=(\,0 \,\,1 \,)^{\text{T}}$ because of the block $1$ in each of their corners.
The moral of the story is that non-commuting observables only imply the OP's first sentence, not the second that I've italicized (see above). Non-commuting observables imply that they can't share an entire common eigenbasis.

Edit
I was wrong about the statement being subtly incorrect (see @Kostya's comments below). The originally intended meaning by the OP (which is how I also first understood it, and which I know was actually the intended meaning by the OP because of their subsequent comments) was incorrect, but the way it was worded actually negated the issue resulting in a technically correct statement. In general an eigenvector of one matrix $A$  will not be an eigenvector of another matrix $B$ when $[A,B]\neq 0$. There may be some eigenvectors of $A$ in common with $B$ (as I originally pointed out), but not all will be in common.
A: Non commuting observables means that a so called measurement is capable of changing the state of the system.
For instance, when there are two observables A and B that fail to commute, there is an eigenvector of A that it is not an eigenvector of B.
When you interact with A then A then B the two results of the A interaction always agree with each other. That means the A interaction always leaves it in a special state, one that gives a specific definite results for an A interaction (the same specific result as the first one gave).
But when that state fails to be an eigenvector of B (and some eigenvector of one fails to be eigen to the other if they fail to commute) then interacting with A then B then A can give two different results for the A interactions.
This proves, definitively, that the interaction with B isn't a passive reveal of preexisting information, but is an interaction that can change the state in question.
Specifically it can change the state from one that gives a particular fixed result for an interaction with A into one capable of giving a different result for an interaction with A.
