Value of the total spin quantum number $s$ in a system of two spin-$1/2$ particles Let it be a system of two spin-$1/2$ particles, the first one in a state $|s_1,m_1 \rangle$ and the second one in a state $|s_2,m_2 \rangle$, with $s_1=s_2=1/2$ and $m_1,m_2=\pm1/2$.
I see that the total secondary spin quantum number $m$ is just the sum of those of each particle, $m=m_1+m_2$, since $\hat S_T|s,m \rangle=(\hat S_1+\hat S_2)(|s_1,m_1 \rangle\otimes|s_2,m_2 \rangle)=\hbar(m_1+m_2)|s,m \rangle$.
However, what happens to the total spin quantum number, $s$? Is it also the sum of $s_1$ and $s_2$? In addition, since $s_1=s_2=1/2\geq0$, how can it be that the so called singlet states have $s=0$: $|s,m\rangle=|0,0\rangle$?
 A: First off, a system composed of two independent spin-$1/2$ particles, with definite spin-projections, would be represented by the tensor product state  $|s_1,m_1\rangle\otimes|s_2,m_2\rangle$, not what you wrote $|s_1,m_1\rangle+|s_2,m_2\rangle$. If you actually try to do the math with what you wrote, you'll see that it's ill-defined - what does it mean to act with $\hat S_{1,z}$ on $|s_2,m_2\rangle$? (Note: often people will omit the $\otimes$ symbol)
Second off, you cannot generically say $|s,m\rangle = |s_1,m_1\rangle\otimes|s_2,m_2\rangle$, because this notation presumes that the LHS is simultaneously an eigenvector of the operators $\hat S^2$ and $\hat S_{z}$ (that's what it means to write the labels $s$ and $m$), but you don't know whether that's true. Indeed, in general it is not. Therefore, we should label this state $|\psi\rangle$, and check whether or not it is an eigenstate of both $\hat S^2$ and $\hat S_{z}$. If it is, then it shall be permissible to label it $|\psi\rangle=|s,m\rangle$, whatever $s$ and $m$ turn out to be.
If you have independently measured the azimuthally-projected spin ($z$-component) of each particle, then indeed the total azimuthally-projected spin will be the sum of each.
$$\begin{align}
\hat S_z|\psi\rangle &= \left(\hat S_{1,z}+\hat S_{2,z}\right)\left(|s_1,m_1\rangle\otimes|s_2,m_2\rangle\right)\\
&= \left(\hbar m_1+\hbar m_2\right)|s_1,m_1\rangle\otimes|s_2,m_2\rangle\\
&=\hbar (m_1+m_2)|\psi\rangle
\end{align}$$
Great! $|\psi\rangle$ is an eigenstate of the operator $\hat S_z$, with eigenvalue $m=m_1+m_2$. Now we check $\hat S^2$.
$$\begin{align}
\hat S^2 |s,m\rangle &= \left(\hat S_1 + \hat S_2\right)^2|s_1,m_1\rangle\otimes|s_2,m_2\rangle\\
&=\left(\hat S_1^2+ \hat S_2^2+2\hat S_1\cdot \hat S_2\right)|s_1,m_1\rangle\otimes|s_2,m_2\rangle\\
&=\left(\frac{3}{4}\hbar^2+ \frac{3}{4}\hbar^2+\underset{?}{\underbrace{2\hat S_1\cdot \hat S_2}}\right)|s_1,m_1\rangle\otimes|s_2,m_2\rangle\\
\end{align}
$$
The math gets a little tricky here. We can write
$$\hat S_1\cdot \hat S_2=\hat S_{1,x} \hat S_{2,x}+\hat S_{1,y} \hat S_{2,y}+\hat S_{1,z} \hat S_{2,z}$$
but the state which we are using, $|s_1,m_1\rangle\otimes |s_2,m_2\rangle$, is not an eigenstate of the $x$ and $y$ spin-projection operators. So we would need to consider each individual case.
In the end, if you do the math, you will find that only the states $|\uparrow \uparrow\rangle$ and $|\downarrow \downarrow\rangle$ are eigenstates of $\hat S^2$, with eigenvalue $2\hbar^2 \implies s=1$.
The remaining states $|\uparrow \downarrow\rangle$ and $| \downarrow\uparrow\rangle$ are not eigenstates, but you can form two independent linear combinations which are eigenstates of $\hat S^2$. These are
$$\begin{align}
\frac{1}{\sqrt{2}}\left[|\uparrow \downarrow\rangle + | \downarrow\uparrow\rangle\right] &\longrightarrow s=1 \\
\frac{1}{\sqrt{2}}\left[|\uparrow \downarrow\rangle - | \downarrow\uparrow\rangle\right] &\longrightarrow s=0 
\end{align}$$
This gets really complicated when dealing with higher spins, or more particles. The formalism for dealing with all this is representation theory of $SU(2)$ (complexification of rotation group). I can explain more about this if you would like.
