Why does the $s$ (and $m$) from the eigenstate $\big |s,m \big >$ come outside of the state and into the eigenvalue along with $\hbar$? In my quantum mechanics class I've learned a notation for basis states which is $\big|s,m\big>$. From what I understand s is the spin (so if a particle was a spin $3\over 2$ it would always be in that position) and m is a projection along that basis which can have values $3\over 2$,$1\over 2$, $-{1\over 2}$, and $-{3\over 2}$. Before we had just been using $\hat J\big|\pm\bf z \big> =\pm{\hbar\over 2} \big| \pm \bf z \big> $. I understand that this isn't useful if you have more than two states it could possibly be. 
Now I'm seeing equations written like this:
$$\mathbf {\hat S}^2\ \big |s,m \big>=s(s+1)\hbar^2\big |s,m \big>$$
However I'm confused by the s coming out of the basis state. Isn't $\big|s,m\big >$ an eigenstate and the value that's in front of it is it's corresponding eigenvalue? Why is s, which is a part of the eigenstate, out in the eigenvalue part of the equation? 
I have the same confusion with this equation as well:
$$\hat S_z\big|s,m\big>=m\hbar \big|s,m\big>$$
What is m doing out there? I have a hunch that $\hat J\big|\pm\bf z \big> ={\hbar\over 2} \big| \pm \bf z \big> $ can also be written like $$\hat J\big|{1\over 2},\pm{1\over 2} \big> =\pm{\hbar\over 2} \big| {1\over 2},\pm{1\over 2} \big> $$ and just like the $\hat S_z$ equation $1\over 2$came out (in the place of m). 
However, this doesn't help me understand why s came out of the eigenstate for ${\bf \hat S}^2$. What is the s doing out there and how is it decided which letter comes out when? I've really just been presented with these formulae without much explanation and I'm having difficulty applying them (especially if I need to change the basis to be along x instead of z).
 A: What's happening is that we are labeling states by their eigenvalues. By definition, the state $|\psi\rangle$ such that $S^2 |\psi\rangle = s(s+1) \hbar^2 |\psi\rangle$ and $S_z |\psi\rangle = m \hbar |\psi\rangle$ will be denoted by $|s, m\rangle$ instead of $|\psi\rangle$. It's just notation, but it's very helpful because we avoid using redundant names like $\psi$: the eigenvalue is all we need to know about the state, so we use it as the name.
You are correct that the states we call $|1/2, \pm 1/2\rangle$ are exactly the same as the $|\pm \hat{\mathbf{z}}\rangle$. In the latter, the fact that the total spin $s$ is $1/2$ is implicit; like you said, if we want to have more general values of $s$ we need to say which is it.
This is part of the reason Dirac notation is so nice, which is that we avoid ugly names like $v_{s,m}$ and instead just use the $|\ \rangle$ to denote that something is a vector, and then put whatever we want inside the bracket. However, we do run into a bit of a problem if we want to use $S_x$ instead of $S_z$ to label states, because the fact that $m$ is the eigenvalue of $S_z$ is implicit. In that case we would either have to warn the reader that $m$ will stand for the eigenvalue of $S_x$ (without the $\hbar$), or use notation like $|s,m_x\rangle$. Again, this is part of the beauty of the whole thing: you can write whatever you want inside the bracket. You will even see people write things like $|\text{alive}\rangle$ and $|\text{dead}\rangle$ when talking about quantum cats.
