Let me remark that this is not the typical setting of Bell inequalities. In a more standard Bell scenario, the two parties/boxes can choose between two different measurement settings, and the violation is observed in the amount of correlation between the observed outcomes (quantified by a specific quantity).
This not to say that the framework discussed in the article is not interesting, but just that I find a bit disingenuous to sell it as a standard Bell scenario (unless this is remarked in the article, which in fairness I haven't read in full).
Where do these 50% come from?
One way to describe the quantum version of this setup is by characterising each measurement choice with a pair of orthonormal vectors/pair of states. Write the state corresponding to the $i$-th outcome for the $j$-th measurement setting on the $k$-th box as $|u_i^{j;k}\rangle$ (the possible values of the indices are $i\in\{1,2\}$, $j\in\{1,2,3\}$, $k\in\{A,B\}$).
Suppose that the measurement operators correspond to the Pauli matrices, which is a canonical set of orthogonal measurement operators. Then, e.g., $|u_i^{1;A}\rangle$ are the eigenstates of the Pauli $X$ operator on the first box. More explicitly,
$$
\begin{array}{cc}
|u_1^{1;A}\rangle=|+\rangle_A\quad & |u_2^{1;A}\rangle=|-\rangle_A \\
|u_1^{2;A}\rangle=|L\rangle_A\quad & |u_2^{2;A}\rangle=|R\rangle_A \\
|u_1^{3;A}\rangle=|0\rangle_A\quad & |u_2^{3;A}\rangle=|1\rangle_A.
\end{array}$$
We can use the same measurement operators/states for the second box (so the corresponding table would be the same modulo $A$ becoming $B$).
We are here using the standard convention to denote the eigenstates of the Pauli matrices: $|\pm\rangle\equiv\frac{1}{\sqrt2}(|0\rangle\pm|1\rangle)$, $|L,R\rangle\equiv\frac{1}{\sqrt2}(|0\rangle\pm i|1\rangle)$.
Ok, so what are then the possible outcomes when we press the same button on both boxes? These are obtained from the overlaps $\langle u_i^{1;A},u_i^{1;B}|\Psi\rangle$, and it is then easy to check that
$$
|\langle u_i^{j;A},u_i^{j;B}|\Psi\rangle|^2 = \frac12,
$$
for all $i=1,2$ and $j=1,2,3$.
For example,
$$\langle u_1^{1;A},u_1^{1;B}|\Psi\rangle = \langle+,+|\Psi\rangle = \frac1{\sqrt2},$$
and the other cases proceed similarly. Note how this implies that
$|\langle u_i^{j;A},u_{1-i}^{j;B}|\Psi\rangle|^2=0$ for all $i,j$, which means that we must always find the same outcome on both boxes when the same measurement is used, consistently with the conditions given by the problem.
What about the probability of getting the same outcome when different measurements are used? These are given by
$$
|\langle u_i^{k;A},u_i^{\ell;B}|\Psi\rangle|^2 = \frac14
$$
for $j\neq k$ and all $i=1,2$. Actually, you can verify that the following more general result holds:
$$
|\langle u_i^{k;A},u_j^{\ell;B}|\Psi\rangle|^2 = \frac14,
$$
for all $i,j$ and $k\neq \ell$. This means that, when different buttons are pressed, the four possible outcomes are all equally probable (whic in particular implies that the probability of getting the same outcome on both boxes is $1/2$).
I suspect, the author means that measuring a property of a quantum particle changing the property, so reading two boolean properties of the same particle is like flipping two coins - with the probability of 12 there will be two heads or two tails.
Well, this turns out to be true (in the sense that the output probabilities do indeed turn out to work like that when different buttons are pressed), but I don't think it's fair to say that this is what the author "means". I don't see any "intuitive" way to see that this happens without passing through some of the math. Quantum mechanics is weird like that.
This contradicts to my impression that measuring the same property twice gives the same result (but kind of resets the other properties). At least, this is how it is explained in Susskind's "Quantum Mechanics: The Theoretical Minimum".
I don't see the contradiction with the previous statement. This is indeed true, measuring the same property twice (at infinitesimally close times) gives the same outcome, but this is not relevant in the kind of setting we are dealing with here. There is an implicit assumption in these kinds of setup that every time you press the buttons the underlying quantum state is somehow "reset" to its initial state.
What, in your opinion, is the best illustration of Bell's theorem, that does not involve too much math but helps to understand what is the theorem about?
I'm afraid I don't know any way to understand the result without passing through some of the math.
