What are $S$ and $L$? In a helium atom, the total spin of the electrons is 0, suggesting that the total spin quantum number S is the sum of the ms quantum numbers (1/2-1/2=0). However, many sources say that the total spin quantum number S is the sum of the s quantum numbers, so in helium S should be 1 (s is always 1/2 for electrons).
In much the same way, I've seen some sources say that L is the sum of the ml quantum numbers and others say that L is the sum of the l quantum numbers.
So, which is correct?
 A: Both $L$ and $S$ are quantum angular momenta. 
Angular momentum is different in quantum mechanics (QM). Almost everything is different in QM.
In QM, angular momentum is complete if you give two numbers: the "angular momentum" and its "third component" $M$.  So you give the pairs  $|l,\ m_l\rangle$ and $|s, \ m_s\rangle$ 
So, any angular momentum is defined with those two numbers. But those numbers behave very differently.
The third component is additive. You add $m$'s without problem:
Total $m$ = $m_1+m_2$
On the other hand, $l$'s behave very differently. When we add two angular momenta, the total angular momentum (let's call it $J$), is different: it can go from $|L-S|$ to $L+S$ 
And the thing is that both things are called "sum", that's the reason of the confusion.
When you add two angular momenta (vectors) $\vec{L}+\vec{S}=\vec{J}$
That's okay, but this is the same as saying
$$|l,\ m_l\rangle \ + \ |s,\ m_s\rangle \ = \ |j,\ m_j\rangle $$
And the total $m$ adds normally:  $m_j=m_l+m_s$, okay.
But $j$ is different. $j$ can go from $|l-s|$ to $|l+s|$ in unit steps.

In all angular momenta, as you know, $m$ can go from $-l$ to $+l$.
And also $m_s$ from $-s$ to $+s$;  and $m_j$ from $-j$ to $+j$.
So, if you have $s=½$, you can have $m_s=\pm ½$
But if you add two spins, $s_1+s_2$, the resulting $j$ can be 0 or 1.
For  $j=0$, $m_j$ can only be 0 as well.
But for $j=1$, $m$ can have three values: $m_j={-1,0,1}$.
So, "summing two spins" can mean that total spin is "J=1" but "M=0".
