The rotation operator normalization you have chosen, $ R_{\boldsymbol n}(\theta)=e^{-i\theta \boldsymbol \sigma\cdot \boldsymbol n/2} $, means that for a rotation by $2\pi$ about the $z$ axis,
$$ R_{\hat z}(2\pi)=e^{-i\pi \sigma_z} =(-1)^{\sigma_z}=-1,$$
because $(-1)^1=(-1)^{-1}=-1$. Thus, your rotation operator has indeed rotated you back to the same point on the Bloch sphere (because the operator is scalar) but you have accumulated a global phase. There isn't, strictly speaking, an error, because global phases can be ignored.
On the other hand, this 'global' phase is tremendously useful if you are in contact with some other degree of freedom. It is an example of a geometric phase (i.e. if you rotate around some path in the Bloch sphere and return to the same path, you will accumulate a phase of
$$\text{(area enclosed by the path in steradians)}/2,$$
where the total area to be had is of course $4\pi$). It is the foundation of what are called phase gates, which are very popular basic entangling gates between two qubits. The idea there is that you perform a controlled rotation: leave qubit A alone if B is in the down state, and rotate it once around the Bloch sphere if B is in the up state. Either way, you come back to the same state, but in one of the two paths you have accumulated a phase, which can be used to generate entanglement.
The Bloch sphere is beautifully minimalist.
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. (This is covered further in the comments below but briefly: QM only predicts averages $\langle \psi|\hat A|\psi\rangle$ and shifting the overall phase of a wave function by some $|\psi\rangle\mapsto e^{i\theta}|\psi\rangle$ therefore cancels itself out in every prediction.)
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. Et 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.
It’s also physical, a real sphere in 3D space.
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. (Here “sort-of” means: it is, if you follow up with an $x$-measurement.)
So let’s ask “which direction is the spin-$\frac12$ most spinning in?” This requires constructing an observable. To give an example, if the $+z$-direction is most-spun-in by a state $|\uparrow\rangle$ then the observable for $z$-spin is the Pauli matrix $\sigma_z = |\uparrow\rangle\langle\uparrow| - |\downarrow\rangle\langle\downarrow|,$ $+1$ in that state, $-1$ in the Hilbert-perpendicular state $\langle \downarrow | \uparrow \rangle = 0.$ Similarly if you look at $\sigma_x = |\uparrow\rangle \langle \downarrow | + |\downarrow \rangle\langle \uparrow |$ you will see that the $|{+x}\rangle$ state defined above is an eigenvector with eigenvalue +1 and similarly there should be a $|{-x}\rangle \propto |\uparrow\rangle - |\downarrow\rangle$ satisfying $\langle {+x}|{-x}\rangle = 0,$ and you can recover $\sigma_x = |{+x}\rangle\langle{+x}| - |{-x}\rangle\langle{-x}|.$
Then 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:$$
\begin{align}
|\psi\rangle\langle\psi| - |\bar\psi\rangle\langle\bar\psi| &= \begin{bmatrix}\alpha\\\beta\end{bmatrix}\begin{bmatrix}\alpha^*&\beta^*\end{bmatrix} - \begin{bmatrix}\beta^*\\-\alpha^*\end{bmatrix} \begin{bmatrix}\beta & -\alpha\end{bmatrix}\\
&=\begin{bmatrix}|\alpha|^2 - |\beta|^2 & 2 \alpha\beta^*\\
2\alpha^*\beta & |\beta|^2 - |\alpha|^2\end{bmatrix}
\end{align}$$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:$$\begin{align} v_z &= \cos^2(\theta/2) - \sin^2(\theta/2) &=&~ \cos \theta,\\
v_x &= 2 \cos(\theta/2)\sin(\theta/2) ~\cos(\phi) &=&~ \sin \theta~\cos\phi, \\
v_y &= 2 \cos(\theta/2)\sin(\theta/2) ~\sin(\phi) &=&~ \sin \theta~\sin\phi.
\end{align}$$So the Bloch prescription uses a $(\theta, \phi)$ which 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! (At least, insofar as a spin-up system spins up.) 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.”
How these answer your questions.
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. Conversely 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 indeed a little confusing 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.
Best Answer
What you need is spherical linear interpolation, which is usually done with quaternions. Well, quaternions can be represented by the pauli matrices. If you don't mind the hand-waviness, a rotation operator about the axis $\vec{s}$ is given by $$R = e^{i \frac{\theta}{2} \vec{s} \cdot \vec{\sigma}}$$.
The vector $\vec{s}$ is perpendicular to the great circle that passes between the two points. You could then find the required angle with a bit of spherical trigonometry.
Edit:
Let $R_e$ be the rotation that takes $\left|0\right\rangle$ to $\left|e\right\rangle$, and similarly $R_g$ takes $\left|0\right\rangle$ to $\left|g\right\rangle$.Then $R_e R_g^\dagger$ takes $\left|g\right\rangle$ to $\left|e\right\rangle$. Suppose I write a qubit as
$$\left|\psi\right\rangle = e^{-i \frac{\phi}{2}} \cos\left(\frac{\theta}{2}\right) \left|0\right\rangle + e^{i \frac{\phi}{2}} \sin\left(\frac{\theta}{2}\right) \left|1\right\rangle$$ I can also write this as $$\left|\psi\right\rangle = \frac{1}{2}\begin{pmatrix} e^{ -i \frac{\phi}{2} } & 0 \\ 0 & e^{ +i \frac{\phi}{2} }\end{pmatrix} \begin{pmatrix} e^{ -i \frac{\theta}{2} } & -e^{ -i \frac{\theta}{2} } \\ e^{ +i \frac{\theta}{2} } & e^{ +i \frac{\theta}{2} }\end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$.
Edit 2: To answer your actual question :) in your case, $$\left|\psi\right\rangle = \frac{1}{2}\begin{pmatrix} e^{ -i \frac{\theta}{2} } & -e^{ -i \frac{\theta}{2} } \\ e^{ +i \frac{\theta}{2} } & e^{ +i \frac{\theta}{2} }\end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ which you can easily write in terms of Pauli matrices.