When we say electron spin is 1/2, what exactly does it mean, 1/2 of what? When we say electron has spin of $\frac{1}{2}$, is that the value of the total spin of electron, or the projection on z axis, or the spin quantum number?
When we say "electron has spin of $\frac{1}{2}\hbar$", is that the value of the total spin or projection? Also, sometimes people say just "spin 1/2" without $\hbar$.
Is spin quantum number $s$ analogous to the l (total orbital angular momentum) or to the $m_s$ (projection of l).
I am confused because when I am trying to learn addition of angular momeneta (eg. in j-j coupling) where we use formula: 
$$\vec{j}=\vec{l}+\vec{s}$$ 
to get total angular momentum for the particle and then we sum all into:
$$\vec{J}=\sum \vec{j}$$ 
What is the $s$ in this context? I mean in the equation: $\vec{j}=\vec{l}+\vec{s}$ since we are summing it with $\vec{l}$ then it must be spin projection on z axis right? 
 A: Given an angular momentum operator with components $S_1, S_2, S_3$ and commutation relations $[S_i, S_j] = \sum_k \epsilon_{ijk}S_k$, where $\epsilon_{ijk}$ are structure constant of the $\mathfrak{su}(2)$ algebra, the Casimir operator $S^2 = S_1^2+S_2^2+S_3^2$ can be diagonalised simultaneously with any of the original components $S_j$ onto their eigenstates $|\psi\rangle$. Furthermore, the below holds:
$$
S^2|\psi\rangle = \hbar^2 s(s+1)|\psi\rangle \qquad S_j|\psi\rangle = \hbar m_j|\psi\rangle.
$$
The value $s$ is said to be the spin of the state, $m_j$ being its projection onto the $j$-direction. According to the structure constants and the Lie-algebras the angular momentum operators close, different values of $s$ are allowed. In the case of electrons we have $s=1/2$. 
A: When we say that the electron has "spin half," we mean half of the quantum of angular momentum, $\hbar$.  A good quantum mechanics text or other reference will help you derive that the Laplacian operator transforms into spherical coordinates like
\begin{align}
\nabla^2 &= 
\left(\frac\partial{\partial x}\right)^2
+
\left(\frac\partial{\partial y}\right)^2
+
\left(\frac\partial{\partial z}\right)^2
\\
&=
\frac 1{r^2}\frac\partial{\partial r}\left(r^2\frac\partial{\partial r}\right)
+
\frac 1{r^2 \sin^2\theta}\left(\frac\partial{\partial\theta}\right)^2
+
\frac 1{r^2 \sin\phi}\frac\partial{\partial\phi}\left(\sin\phi\frac\partial{\partial\theta}\right)
\end{align}
The angular parts of this operator act on the spherical harmonics to give eigenvalue $\ell(\ell+1)$ for integer $\ell$.
This means that the effective form of the kinetic energy operator is
\begin{align}
\frac{\hbar^2}{2m}\nabla^2
&=
\frac{\hbar^2}{2mr^2}\frac\partial{\partial r}\left(r^2\frac\partial{\partial r}\right)
+
\boxed{\frac{\hbar^2}{2mr^2}{\ell(\ell+1)}}
\end{align}
In the limit of large $\ell$, the term in the box is the same as the orbital kinetic energy for a point mass $m$ rotating some $r$ from the center of motion with angular momentum $L\sim\ell\hbar$.
This argument is what lets us say things like "$\hbar$ is the quantum of angular momentum," or "angular momentum comes in lumps, and the size of each lump is $\hbar$."  Since $\hbar$ is the only quantum of angular momentum, sometimes we only count quanta and leave the unit off.  Same as when someone quotes you a price and gives the value but not the currency ("I'll take your car off this tow truck for fifty-five").
Spin angular momentum falls naturally out of the Dirac question in a surprisingly elegant way.  You get the same quantum, $\hbar$.  However the Dirac equation describes objects whose intrinsic angular momentum is $\hbar/2$.
Therefore the projection $m_s$ of the electron spin along any axis can be $\pm\frac12\hbar$, but never zero.
I think this might clarify your search for guidance on the rules for summing vector angular momenta.
A: The explanation is very simple.  Based on the Stern- Gerlach experiment spin of 1/2 simply means that if you fire electrons through his apparatus... 1/2 of the electrons will spin up and the other 1/2 will spin down
A: To explain simply, without getting into the details of rotation symmetry groups, etc:
When one says $s=1/2$, or $m_s=1/2$ or $m_s=-1/2$, we are specifying a quantum number which describes how the eigenvalues of spin operators behave.
If we specify an eigenket of the spin angular momentum operators $|s,m_s\rangle = |1/2, \pm 1/2\rangle$, with operators $\hat{s^2}$ and $\hat{s_z}$ then
$$\hat{s^2} |1/2, \pm 1/2\rangle = \frac{3}{4} \hbar^2~ |1/2, \pm 1/2\rangle \\
\hat{s_z} |1/2, \pm 1/2\rangle = \pm\frac{1}{2} \hbar~ |1/2, \pm 1/2\rangle
$$
When one is combining quantized angular momenta, there are rules from the symmetry groups that help us determine the allowed quantum numbers. The allowed quantum numbers follow a triangle rule. Suppose we want to find the allowed quantum numbers for a state resulting from the combination of two angular momenta $|\ell_1, m_{\ell 1}\rangle$ and $|\ell_2, m_{\ell 2}\rangle$:
$$ \ell_1 +\ell_2 \ge \ell_{new} \ge |\ell_1 - \ell_2|$$ with integer steps between the minimum and maximum, and z-component quantum number merely sum:
$$  m_{\ell \mathrm{new}}=m_{\ell 1}+m_{\ell 2}$$.
Where $\ell$ represents any type of angular momentum quantum number (spin, orbital, spin-orbit combo, etc).
