How does using a Bell state lead to a $\cos \left(\frac{1}{8}\pi \right)$ probability of winning in the CHSH game? I have trouble understanding how the CHSH (which stands for John Clauser, Michael Horne, Abner Shimony, and Richard Holt) game, as described in this paper (and shortly explained in this post), works. 
I understand that $75\%$ is the maximum probability of winning in a classical system.  
The following Bell state
$$\frac{\left| 00 \right> + \left| 11 \right>}{\sqrt{2}}$$
can be interpreted as having $50\%$ chance that both qubits are $\left| 0\right>$ and $50\%$ chance that they are both $\left| 1 \right>$. 
This state can be prepared using the following gate

However, it is unclear to me how the Bell state (above) leads to a $\cos\left(\frac{1}{8}\pi\right)\approx0.85$ probability of winning in the CHSH game.
I have made a visual representation of the Bloch sphere from one side in Desmos, and I see how a certain angle corresponds to a certain probability. This might be an incorrect interpretation, but this is how I picture a qubit.  
So, how does the CHSH game conclude that the probability to 'win' in a quantum system is $\cos(\frac{1}{8}\pi)$, or an angle of $45°$ in my Desmos example?
 A: $\newcommand{\ket}[1]{\left|#1\right>}$
Indeed, the story about $\cos\frac{\pi}{8}$, is not told often, except in quantum mechanics lesson, and even there, it is often left as “exercise left to the reader”. This number is not obvious at first sight, but is the result of an optimization and a straightforward application of quantum mechanics computation rules.
What is apparently lacking in your description to find it is the description of the measurements. 
To keep things (relatively) simple, I will assume that we deal with single photons entangled in polarization, and I will only consider linear polarization. Let denote by $α$ (rep. $β$) the angle defining Alice’s (resp. Bob’s) measurement. Measuring the polarization of a single photon in a direction $α$, is a binary measurement, giving $0$ if the photon is oriented along $α$, and $1$ if it is oriented along $α+\frac{π}{2}$. Measuring along this direction, is equivalent to first rotate the photon by an angle $-α$, ant then measure it in the vertical-horizontal (a.k.a $α=0$) basis. This rotation is a linear transformation, transforming Alice’s state as follows:
$$\ket{0}:↦\cosα\ket0 - \sinα \ket1$$
$$\ket{1}:↦\sinα\ket0 + \cosα \ket1$$
Bob’s states transform in a similar way, leading, for the global state, to
$$
\frac{\ket{00}+\ket{11}}{\sqrt2}:↦\\(\cosα\cosβ+\sinα\sinβ)\frac{\ket{00}+\ket{11}}{\sqrt2}+(-\cosα\sinβ+\sinα\cosβ)\frac{\ket{01}-\ket{10}}{\sqrt2}$$
To find the optimal measurement for the CHSH game, you need then to optimize over the possible sets of angles $(α,α',β,β')$. Of course, being clever and knowing trigonometric formulæ helps in this optimization. (Knowing that the answer is $(0,\frac{π}{4},\frac{π}{8},\frac{3π}{8})$ helps too !).
Following the habits of the field, I leave the complete computation as an exercise to the reader ;-).
By the way, this only shows that the CHSH game can be won with 85% success rate with quantum entanglement. The fact that one cannot do better  is known as Tsirelson’s bound, and 
involves linear algebra.
Footnotes
¹: *If it where more obvious, many discussions on the nature of entanglement might have happened much before the 1960’s *
