How is quantization related to commutation? How are commutation (of observables) and quantization related? Reading about the Stone-Von Neumann Theorem, it seems that commutativity is the classical limit of quantum mechanics, and hence non-quantization, but I don't understand the intuition behind the fact that commutativity of operators should imply any kind of quantization.
In a theory involving `quantized' quantities, such as quantum mechanics, why does commutation of operators (observables) suddenly become an important topic- why does quantization come hand in hand with the uncertainty principle? 
 A: I believe that you must study (not simply reading but trying to prove) that beautiful story of angular momentum quantization starting from the non-commutativity of the coordinate and momentum of a particle along any axis. The following is a starting point (found in any introductory book on Quantum Mechanics).     
In Classical Mechanics the angular momentum of a particle $\mathbf{J}=\left(J_1,J_2,J_3\right)$ is defined by
\begin{equation}
   \mathbf{J}\  \equiv \ \mathbf{r}\times\mathbf{p}\ = 
   \begin{vmatrix}
         &\mathbf{e}_1&\mathbf{e}_2&\mathbf{e}_3&\\
         &x_1&x_2&x_3&\\
         &p_1&p_2&p_3&
       \end{vmatrix}
   \tag{01}
\end{equation}
where $ \mathbf{r}=\left(x_1,x_2,x_3\right)$ the position, $\mathbf{p}=\left(p_1,p_2,p_3\right)$ the momentum of the particle and $\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \}$ the basic vectors  of an orthonormal Cartesian coordinate system.
The definition (01) is valid in Quantum Mechanics too, but the $x_j,p_k$ are hermitian operators. So in coordinates
\begin{equation}
   \mathbf{J}=
      \begin{bmatrix}
         &J_1&\\
         &J_2&\\
         &J_3&
       \end{bmatrix}
   \equiv
      \begin{bmatrix}
         &x_2p_3-x_3p_2&\\
         &x_3p_1-x_1p_3&\\
         &x_1p_2-x_2p_1&
       \end{bmatrix}  
   =\ \mathbf{r}\times\mathbf{p}   
   \tag{02}
\end{equation}
Now, we'll try to find if the components $J_1,\ J_2, \ J_3$ commute between each other or not, that is, if we can measure two components precisely at the same time.
In order to find the commutators $[J_j,J_k],\ j\neq k$, we remind the commutation relations of $x_j,p_k$
\begin{eqnarray}
   &\left[x_j,p_k\right] &=\  x_jp_k-p_kx_j\ = \ i \hbar \delta_{jk}
   \tag{03-1}\\
   &\left[x_j,x_k\right] &=\  x_jx_k-x_kx_j\ =\ 0
   \tag{03-2}\\
   &\left[p_j,p_k\right] &=\ p_jp_k-p_kp_j\ =\ 0
   \tag{03-3} 
\end{eqnarray}
So
\begin{eqnarray}
   \left[J_1,J_2\right] &=& J_1J_2-J_2J_1 \nonumber\\   &=&\left(x_2p_3-x_3p_2\right)\left(x_3p_1-x_1p_3\right)-\left(x_3p_1-x_1p_3\right)\left(x_2p_3-x_3p_2\right) \nonumber \\
   &=&\left(x_1p_2-x_2p_1\right)\left(x_3p_3-p_3x_3\right)\nonumber
\end{eqnarray}
or
\begin{equation}
   \left[J_1,J_2\right]\  =\  J_1J_2-J_2J_1 \ =\  i \hbar J_3\\
   \tag{04-1}   
\end{equation}
and in a similar way for the other two commutators
\begin{eqnarray}
   \left[J_2,J_3\right]&=& J_2J_3-J_3J_2 \ =\  i \hbar J_1
   \tag{04-2}\\
   \left[J_3,J_1\right]&=& J_3J_1-J_1J_3 \ =\  i \hbar J_2
   \tag{04-3}
\end{eqnarray}
Equations  (04) are expressed together under the symbolic and more simple relation
\begin{equation}
   \mathbf{J}\times \mathbf{J}\  = \ i\;\hbar \;\mathbf{J}\\
   \tag{05}   
\end{equation}
or using the dimensionless operator
\begin{equation}
     \mathbf{L}=\left(L_1,L_2,L_3\right)\equiv \left(\frac{J_1}{\hbar},\frac{J_2}{\hbar},\frac{J_3}{\hbar}\right)=\frac{\mathbf{J}}{\hbar}\\
   \tag{06}    
\end{equation}
\begin{equation}
   \mathbf{L}\times \mathbf{L}\  = \ i\;\mathbf{L}\\
   \tag{07}   
\end{equation}
It's impossible to measure precisely at the same time any two components of the angular momentum since they don't commute.
The story continues by the fact that the operator
\begin{equation}
   L^2\equiv L_1^2+L_2^2+L_3^2
   \tag{08}    
\end{equation}
which represents the square of the absolute value of angular momentum $\mathbf{L}$, commutes with its component along any axis, say $\:L_3 \:$
\begin{equation}
\left[L^2,L_3\right]=L^2L_3-L_3L^2=0
\tag{09}
\end{equation}
and in general
\begin{equation}
\qquad \qquad \qquad \quad \left[L^2,L_k\right]=0 ,\qquad k=1,2,3
\tag{10}
\end{equation}
We can measure precisely at the same time the absolute value of angular momentum $L^2$ and its component $L_k$ along any arbitrary axis $x_k$.
The finale is

THEOREM :
The components of the dimensionless orbital angular momentum of a particle $\:\mathbf{L}=\mathbf{J}/\hbar =\left(\mathbf{r}\times\mathbf{p}\right)/ \hbar $ satisfy the commutation relations $\mathbf{L}\times \mathbf{L}= i\mathbf{L}$, by which  
a) The allowable eigenvalues of the absolute value operator
\begin{equation}  
 L^2\equiv L_1^2+L_2^2+L_3^2
  \nonumber\\
\end{equation}
are
\begin{equation}   
   \bbox[#FFFF88,5px]{j\left(j+1\right)\;, \qquad j=0\:,\:\tfrac{1}{2}\:,\:1\:,\:\tfrac{3}{2}\:,\:2\:,\:\tfrac{5}{2}\:,\:\ldots}
   \nonumber\\
\end{equation}
b) The eigenvalue $\ j\left(j+1\right)\ $ has $\ (2j+1)$-multiplicity 
to which correspond the $\ (2j+1)\ $ possible eigenvalues of the component $L_3$ across an arbitrary axis $x_3$
\begin{equation}
\bbox[#E6E6E6,5px]{m\;, \qquad   m = -j\:,\:-j+1\:,\:\cdots \:,\:j-1\:,\:j}
   \nonumber\\
\end{equation}
c) A complete set of common eigenfunctions of $L^2$ and $L_3$ is enumerated by the pair $(j,m)$.  
(Note : it has been proved that the orbital angular momentum  has only integer values of $j$ while the half-integer values are due exclusively to spin. Moreover, spin can take integer values too).

The non-commutativity of the position and momentum operators along any axis yields the non-commutativity of the components of angular momentum which in turn gives its quantization. This same position-momentum non-commutativity is responsible for the quantization of the energy of the linear harmonic oscillator.  
I don't think that these results are a matter of intuition.
