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.
