It is usually said that the points on the surface of the Bloch sphere represent the pure states of a single 2-level quantum system. A pure state being of the form: $$ |\psi\rangle = a |0\rangle+b |1\rangle $$ And typically the north and south poles of this sphere correspond to the $|0\rangle$ and $|1\rangle$ states. Image: ("Bloch Sphere" by Glosser.ca - Own work. Licensed under CC BY-SA 3.0 via Commons - https://commons.wikimedia.org/wiki/File:Bloch_Sphere.svg#/media/File:Bloch_Sphere.svg) 
But isn't this very confusing? If the north and south poles are chosen, then both states are on the same line and not orthogonal anymore, so how can one choose an arbitrary point $p$ on the surface of the sphere and possibly decompose it in terms of $0,1$ states in order to find $a$ and $b$? Does this mean that one shouldn't regard the Bloch sphere as a valid basis for our system and that it's just a visualization aid?
I have seen decompositions in terms of the internal angles of the sphere, in the form of: $a=\cos{\theta/2}$ and $b=e^{i\phi}\sin{\theta/2}$ with $\theta$ the polar angle and $\phi$ the azimuthal angle. But I am clueless as to how these are obtained when $0,1$ states are on the same line.
Answer
Conventionally, a qubit has four real parameters; $a e^{i\chi} |0\rangle + b e^{i\phi} |1\rangle.$ However, some quick insight reveals that the $a$-vs-$b$ tradeoff only has one degree of freedom due to the normalization $a^2 + b^2 = 1$ and some more careful insight reveals that, in the way we construct expectation values in QM, you cannot observe $\chi$ or $\phi$ themselves but only the difference $\chi - \phi$, which is $2\pi$-periodic.
So if you think at the most abstract about what you need, you just draw a line from 0 to 1 representing the $a$-vs-$b$ tradeoff: how much is this in one of these two states? Then you draw circles around it: how much is the phase difference? What stops it from being a cylinder is that the phase difference ceases to matter when $a=1$ or $b=1$, hence the circles must shrink down to points. And voila, you have something which is topologically equivalent to a sphere. The sphere contains all of the information you need for experiments, and nothing else.
This is the more shocking fact. Given only the simple picture above, you could be forgiven for thinking that this was all harmless mathematics: no! In fact the quintessential qubit is a spin-$\frac 12$ system, with the Pauli matrices indicating the way that the system is spinning around the $x$, $y$, or $z$ axes. This is a system where we identify $|0\rangle$ with $|\uparrow\rangle$, $|1\rangle$ with $|\downarrow\rangle$, and the phase difference comes in by choosing the $+x$-axis via $|+x\rangle = \sqrt{\frac 12} |0\rangle + \sqrt{\frac 12} |1\rangle.$
The orthogonal directions of space are not Hilbert-orthogonal in the QM treatment, because that's just not how the physics of this system works. Hilbert-orthogonal states are incommensurate: if you're in this state, you're definitely not in that one. But this system has a spin with a definite total magnitude of $\sqrt{\langle L^2 \rangle} = \sqrt{3/4} \hbar$, but only $\hbar/2$ of it points in the direction that it is "most pointed along", meaning that it must be distributed on some sort of "ring" around that direction. Accordingly, when you measure that it's in the $+z$-direction it turns out that it's also sort of half in the $+x$, half in the $-x$ direction ("sort of" = it is, if you follow up with an $x$-measurement).
The state orthogonal to $|\psi\rangle = \alpha |0\rangle + \beta |1\rangle$ is $|\bar \psi\rangle = \beta^*|0\rangle - \alpha^* |1\rangle,$ so the observable which is +1 in that state or -1 in the opposite state is:$$|\psi\rangle\langle\psi| - |\bar\psi\rangle\langle\bar\psi| = \begin{bmatrix}\alpha\\\beta\end{bmatrix}\begin{bmatrix}\alpha^*&\beta^*\end{bmatrix} - \begin{bmatrix}\beta^*\\-\alpha^*\end{bmatrix} \begin{bmatrix}\beta & -\alpha\end{bmatrix}=\begin{bmatrix}|\alpha|^2 - |\beta|^2 & 2 \alpha\beta^*\\ 2\alpha^*\beta & |\beta|^2 - |\alpha|^2\end{bmatrix}$$Writing this as $v_i \sigma_i$ where the $\sigma_i$ are the Pauli matrices we get:$$v_z = |\alpha|^2 - |\beta|^2,\\ v_x + i v_y = 2 \alpha^* \beta.$$ Now letting $\alpha = \cos(\theta/2)$ and $\beta = \sin(\theta/2) e^{i\phi}$ we find out that these are:$$\begin{align} v_z &= \cos^2(\theta/2) - \sin^2(\theta/2) &=&~ \cos \theta,\\ v_x &= 2 \cos(\theta/2)\sin(\theta/2) ~\cos(\phi) &=&~ \sin \theta~\cos\phi, \\ v_y &= 2 \cos(\theta/2)\sin(\theta/2) ~\sin(\phi) &=&~ \sin \theta~\sin\phi. \end{align}$$These are simply the spherical coordinates of the point on the sphere which such a $|\psi\rangle$ is "most spinning in the direction of."
So instead of being a purely theoretical visualization, we can say that the spin-$\frac 12$ system, the prototypical qubit, actually spins in the direction given by the Bloch sphere coordinates! It is ruthlessly physical: you want to wave it away into a mathematical corner and it says, "no, for real systems I'm pointed in this direction in real 3D space and you have to pay attention to me."
Yes, N and S are spatially parallel but in the Hilbert space they are orthogonal. This Hilbert-orthogonality means that a system cannot be both spin-up and spin-down. Indeed, the lack of Hilbert-orthogonality between, say, the $z$ and $x$ directions means that when you measure the $z$-spin you can still have nonzero measurements of the spin in the $x$-direction, which is a key feature of such systems. It is probably a little confusing, indeed, to have two different notions of "orthogonal," one for physical space and one for the Hilbert space, but it comes from having two different spaces that you're looking at.
One way to see why the angles are physically very useful is given above. But as mentioned in the first section, you can also view it as a purely mathematical exercise of trying to describe the configuration space with a sphere: then you naturally have the polar angle as the phase difference, which is $2\pi$-periodic, so that is a naturally "azimuthal" coordinate; therefore the way that the coordinate lies along 0/1 should be a 'polar' coordinate with $0$ mapping to $|0\rangle$ and $\pi$ mapping to $|1\rangle$. The obvious way to do this is with $\cos(\theta/2)$ mapping from 1 to 0 along this range, as the amplitude for the $|0\rangle$ state; the fact that $\cos^2 + \sin^2 = 1$ means that the $|1\rangle$ state must pick up a $\sin(\theta/2)$ amplitude to match it.
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