When we say electron spin is 1/2, what exactly does it mean, 1/2 of what?

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.

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 $s=1/2$.