... ball.¹

If you want to be really concrete, imagine a spinning gyroscope fitting snugly in a box. Rotate the box.

... (frames).²

In other words, a element of $SO(3)$ determines a mapping of ${\bf R}^3$ to ${\bf R}^3$. As it happens, the action is faithful, i.e., the mapping determines the element of $SO(3)$.

... moment.³

A quick review: write $(a:b)$ for the equivalence class of $(a,b)$. We will associate either a complex number $w$ or else $\infty$ with each class $(a:b)$. If $b\neq 0$, then $(a:b)=(w:1)$, where $w=a/b$. All pairs of the form $(a,0)$ belong to the class $(1:0)$. So we associate $a/b$ with $(a:b)$ if $b\neq 0$, and $\infty$ with $(1:0)$. Mapping the complex plane plus $\infty$ to the Riemann sphere via the usual stereographic projection completes the trick. Some sample points to bear in mind: the north pole is $(1:0)$; the south pole is $(0:1)$; points on the equator have the form $(e^{i\theta}:1)$. (For the purist, the special treatment of $\infty$ rankles. It is not singled out on either the complex projective line or on the Riemann sphere. Later I will show how to set up the correspondence without this blemish.)

... simple⁴

In the infinite dimensional case, you have to use the spectral decomposition of $A$ instead of an orthonormal basis of eigenvectors; another reason why spin is simpler than position.

... proportional to $\cos\alpha$. ⁵

Why wouldn't the electron simply snap into alignment with the magnetic field? Answer: the spinning electron would act like a gyroscope, and precess in response to the torque exerted by the field. Thus it would maintain its angle of inclination to the field.