Total spin of system of two spin-$1/2$ particles Consider a quantum system of two spin one half particles. Let $\alpha(1)$ be 'spin up' for first system, and $\beta(1)$ 'spin down' for first system, and likewise for second system. We have $$ \chi = \frac{1}{\sqrt{2}} \bigg( \alpha(1) \beta(2) - \beta(1) \alpha(2) \bigg). $$ If I want to compute the total spin of the system, do I just write $$S = S_1 + S_2 = (S_{1x} + S_{1y} + S_{1z}) + (S_{2x} + S_{2y} + S_{2z} ) $$ and act with this operator on $\chi$? Then the components of $S_1$ would act only on state $1$, and the components of $S_2$ on state $2$. Is that the correct approach to compute the total spin? 
 A: There are two quantum numbers to consider, the total spin = 1/2 and the azimuthal or projection quantum number on an axis, say z with angular momentum +-hbar/2. This quantum number is also 1/2 for electrons. (The x and y components are undefined by the uncertainty principle as spin quantum number and spin z component are defined) Thus for 2 electrons there are 4 states produced, one with electron spin paired (total spin =0) which a singlet state with and three triplet states with total spin 1.   In the absence of any external fields these have the, same energy. 
The figure below gives more details.

If you look at some text books on atomic spectroscopy you will find this described in detail. The simplest description is in 'Modern Molecular Photochemistry' by N. Turro, (Chapter 2) more detailed are 'Modern Spectroscopy' by Hollas; (chapter on electronic spectroscopy); 'Molecular Quantum Mechanics' by Atkins & Friedman,(Chapter 4) and 'Molecules & Radiation' by J. Steinfeld. (Chapter 2). Look for Russell Saunders Coupling for an explanation of adding angular momentum in general.
A: In quantum mechanics you cannot "compute the total spin" of a certain quantum state: all you can compute are the expected values of the spin components.
(I'm going to use Dirac notation because I find it much clearer) To compute expected values in your case it is convenient to use the $\mid s,m_z\rangle$ representation. If $\vec S = \vec S_1 + \vec S_2$ is the total spin of the system, you have
$$S^2 \mid s,m_z\rangle =(\vec S_1 + \vec S_2)^2 \mid s,m_z\rangle = s(s+1) \hbar^2 \mid s,m_z\rangle$$
and
$$ S_z  \mid s,m_z\rangle = (S_{z,1}+S_{z,2})  \mid s,m_z\rangle  =  m_z \hbar \mid s,m_z\rangle  $$
The state you want to consider is the singlet state, which we will write in the $\mid s,m_z\rangle$ representation (see here for example):
$$ \mid \chi \rangle = \frac 1 {\sqrt{2}} (\mid \uparrow,\downarrow \rangle - \mid \downarrow,\uparrow \rangle) = \ \mid s=0,m_z=0\rangle \ = \ \mid 0,0 \rangle$$
So you will have:
$$\langle 0,0\mid S^2 \mid 0,0 \rangle = 0$$
$$\langle 0,0\mid S_z \mid 0,0 \rangle = 0$$
To calculate $\langle 0,0\mid S_x \mid 0,0 \rangle$ and $\langle 0,0\mid S_y \mid 0,0 \rangle$, use the fact that
$$S_x=\frac{S_+ + S_-}{2}$$
$$S_y = \frac{S_+-S_-}{2i}$$
where
$$S_{\pm}\mid s,m_z\rangle = \hbar \sqrt{s(s+1)-m_z(m_z\pm1)} \mid s,m_z \pm 1\rangle$$
