In classical mechanics, the angular momentum of a body is given by
$$L=r\times p$$ where $r$ and $p$ denote radius arm and linear momentum respectively. In quantum mechanics, the angular momentum of a spinning particle can be obtained by replacing linear momentum $p$ by momentum operator $-i\hbar\nabla$. As a result, the components of quantum mechanical angular momentum $L$ is given by
\begin{align*}
L_x&=-i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\\
L_y&=-i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\\
L_z&=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)
\end{align*}

It is interesting to see that the quantum mechanical angular momentum can be obtained purely from algebra, more specifically from representation theory. The relevant representations are the representations of the special unitary group $\mathrm{SU}(2)$ and its Lie algebra $\mathfrak{su}(2)$. This hints us that the symmetry of the background space plays a crucial role in quantum mechanics.

The rotations in $\mathbb{R}^3$ form the special orthogonal group $\mathrm{SO}(3)$. $\mathrm{SO}(3)$ is not simply connected (see [3] for detalis) and the special unitary group $\mathrm{SU}(2)$ is the universal covering group of $\mathrm{SO}(3)$. ($\mathrm{SU}(2)=S^3$ so it is simply connected.) The covering map $\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$ is a $\mathrm{SU}(2)$ representation. To see this, note that Euclidean 3-space $\mathbb{R}^3$ can be identified with the set of $2\times 2$ hermitian matrices of the form
$$\underline{X}=\begin{pmatrix}
z & x-iy\\
x+iy & -z
\end{pmatrix}$$
with the inner product $\langle\ ,\ \rangle$ defined by
$$\langle\underline{X},\underline{Y}\rangle=\frac{1}{2}\mathrm{tr}(\underline{X}\cdot\underline{Y})$$
In particular
$$|\underline{X}|^2=\frac{1}{2}\mathrm{tr}\underline{X}^2=-\det\underline{X}$$
Here the hermitian matrix $\underline{X}$ is identified with the vector $(x,y,z)\in\mathbb{R}^3$. Since $|X|^2=-\det\underline{X}$, the identification is an isometry. $\mathrm{SU}(2)$ acts on $\mathbb{R}^3$ isometrically by the action
$$\mathrm{SU}(2)\times\mathbb{R}^3\longrightarrow\mathbb{R}^3;\ (U,X)\longmapsto U^{-1}XU$$
For a fixed $U\in\mathrm{SU}(2)$, the map
$$\mathbb{R}^3\longrightarrow\mathbb{R}^3;\ X\longmapsto U^{-1}XU$$
is an orientation preserving isometry of $\mathbb{R}^3$. Thus the Lie group action induces a Lie group representation $\rho:\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$. Since both $U$ and $-U$ result the same isometry, the representation $\rho$ is a 2:1 map. The kernal of $\rho$ is $\mathbb{Z}_2=\{\pm I\}$, so we have $\mathrm{SU}(2)/\mathbb{Z}_2=\mathrm{SO}(3)$. The quotient group $\mathrm{SU}(2)/\mathbb{Z}_2$ is denoted by $\mathrm{PSU}(2)$ is called the projective special unitary group. The double cover of the special orthogonal group $\mathrm{SO}(n)$ is called the spin group and is denoted by $\mathrm{Spin}(n)$. Hence, the double cover $\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$ is the spin group $\mathrm{Spin}(3)$.

Let $\mathcal{H}$ be the Hilbert space of states $\psi$ as smooth functions on $\mathbb{R}^3$. Define a map $\Pi:\mathrm{SU(2)}\longrightarrow\mathrm{GL(\mathcal{H})}$ as follows: For each $U\in\mathrm{SU}(2)$, $\Pi(U):\mathcal{H}\longrightarrow\mathcal{H}$ is an isomorphism defined by
$$[\Pi(U)\psi](v)=\psi(\rho(U)^{-1}v),\ v\in\mathbb{R}^3$$
where $\rho$ is the universal covering map $\rho: \mathrm{SU}(2)\stackrel{2:1}{\longrightarrow}\mathrm{SO}(3)$. $\Pi$ is indeed a group homomorphism: For $U_1,U_2\in\mathrm{SU}(2)$,
\begin{align*}
\Pi(U_1)[\Pi(U_2)\psi](v)&=\Pi(U_2\psi)(\rho(U_1)^{-1}v)\\
&=\psi(\rho(U_2)^{-1}\rho(U_1)^{-1}v)\\
&=\psi((\rho(U_1)\rho(U_2))^{-1}v)\\
&=\psi(\rho(U_1U_2)^{-1}v)\\
&=[\Pi(U_1U_2)\psi](v)
\end{align*}
Hence, $\Pi$ is an infinite dimensional real representation of $\mathrm{SU}(2)$. Here the fact that $\rho$ is a group homomorphism is used. We can also obtain the corresponding representation $\pi$ of the Lie algebra $\mathfrak{su}(2)$. $\pi$ can be computed as
$$\pi(X)=\frac{d}{dt}\Pi(e^{tX})|_{t=0}$$
So,
\begin{align*}
[\pi(X)\psi](v)&=\frac{d}{dt}[\Pi(e^{tX})\psi](v)|_{t=0}\\
&=\frac{d}{dt}\psi(\rho(e^{tX})^{-1}v)|_{t=0}
\end{align*}
The Lie algebra $\mathfrak{su}(2)$ has the canonical basis
\begin{align*}
X_1&=\frac{i}{2}\sigma_1=\frac{i}{2}\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}\\
X_2&=\frac{i}{2}\sigma_2=\frac{i}{2}\begin{pmatrix}
0 & i\\
-i & 0
\end{pmatrix}\\
X_3&=\frac{i}{2}\sigma_3=\frac{i}{2}\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}
\end{align*}
Let us calculate $\pi$ for the basis member $X_3$. $e^{\theta X_3}=\begin{pmatrix}
e^{i\theta/2} & 0\\
0 & e^{-i\theta/2}
\end{pmatrix}$ and $\rho(e^{\theta X_3})=R_\theta^z$ where $R_\theta^z=\begin{pmatrix}
\cos\theta & -\sin\theta & 0\\
\sin\theta & \cos\theta & 0\\
0 & 0 & 1\end{pmatrix}$ is rotation in $\mathbb{R}^3$ about the $z$-axis by angle $\theta$. Let $v(\theta)$ be a curve in $\mathbb{R}^3$ defined by
$$v(\theta)=\rho(e^{\theta X_3})^{-1}v=(R_\theta^z)^{-1}v$$ so that $v(0)=v$. Write $v(\theta)=(x(\theta),y(\theta),z(\theta))$ and $v=(x,y,z)$. Then by the chain rule,
\begin{align*}
[\pi(X_3)\psi](v)&=\frac{\partial\psi}{\partial x}\frac{dx}{d\theta}|_{\theta=0}+\frac{\partial\psi}{\partial y}\frac{dy}{d\theta}|_{\theta=0}+\frac{\partial\psi}{\partial z}\frac{dz}{d\theta}|_{\theta=0}\\
&=y\frac{\partial\psi}{\partial x}-x\frac{\partial\psi}{\partial y}
\end{align*}
Hence,
$$\pi(X_3)=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}$$
Using
\begin{align*}
e^{\phi X_2}&=\begin{pmatrix}
\cos\frac{\phi}{2} & -\sin\frac{\phi}{2}\\
\sin\frac{\phi}{2} & \cos\frac{\phi}{2}
\end{pmatrix},\ \rho(e^{\phi X_2})=R_\phi^y=\begin{pmatrix}
\cos\phi & 0 & -\sin\phi\\
0 & 1 & 0\\
\sin\phi & 0 & \cos\phi
\end{pmatrix}\\
e^{\sigma X_1}&=\begin{pmatrix}
\cos\frac{\sigma}{2} & i\sin\frac{\sigma}{2}\\
i\sin\frac{\sigma}{2} & \cos\frac{\sigma}{2}
\end{pmatrix},\ \rho(e^{\sigma X_1})=R_\sigma^x=\begin{pmatrix}
1 & 0 & 0\\
0 & \cos\sigma & -\sin\sigma\\
0 & \sin\sigma & \cos\sigma
\end{pmatrix}
\end{align*}
one can find similar formulas for $\pi(X_2)$ and $\pi(X_1)$:
\begin{align*}
\pi(X_2)&=z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\\
\pi(X_1)&=z\frac{\partial}{\partial y}-y\frac{\partial}{\partial z}
\end{align*}
Note that
\begin{align*}
i\hbar\pi(X_1)&=L_x=-i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\\
-i\hbar\pi(X_2)&=L_y=-i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\\
i\hbar\pi(X_3)&=L_z=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)
\end{align*}
i.e. the angular momenta about the $x$-axis, $y$-axis and $z$-axis respectively.

References:

[1] Walter Greiner, Quantum Mechanics, An Introduction, 4th Edition, Springer-Verlag 2000

[2] Brian C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Springer-Verlag 2004

[3] Shlomo Sternberg, Group Theory and Physics, Cambridge University Press 1994