How to compute the regression of individual fitness on individual phenotype
I am not sure I understood the question. Let me know if this helps.

Case: N=2, freq=0.5

Let's assume that the frequency of those who cooperate is 0.5. The slope of the regression line (which R.squared is equal to 1 as we have as many data point than degrees of freedom) is by definition $\frac{\Delta w}{\Delta p}$. You defined $\Delta p = 1$. What is $\Delta w$ then?

Well, given that one cooperate and the other defect (as indicated in the fact that the frequency of cooperators is 0.5), then one will have the payoff of a cooperator that is facing a defector and the other one will have the payoff of a defector facing a cooperator. The cooperator has a fitness of $w_0 + b$ and the defector has a fitness of $w_0 - c$. I think you may misread the payoff matrix. To know the fitness of someone with strategy $i$ when facing someone with strategy $j$, you look that payoff at row $i$ and column $j$.

Therefore, $\Delta w = (w_0 - c) - (w_0 + b) = - (b+c) = -b - c$, and so the slope is  $\frac{\Delta w}{\Delta p}=\frac{-b-c}{1} = - b -c$

Case: N>2, freq=0.5

In such case the slope of the regression will necessarily be lower than $-b-c$ because some cooperators will meet other cooperators, increasing their fitness (=highering the points on the right hand side of your graph) and some defectors will meet other defectors, decreasing their fitness (=lowering the points on the left hand side of your graph)

At equilibrium

The only stable equilibrium is when the frequency of cooperators is 0. In such case, there is no variance and we cannot even talk about regression. 

Conclusion

The slope is not necessarily $-b-c$. It is $-b-c$ for the very special case where we have two individuals, one cooperate, the other one defects. In any other scenario the slope $s$ of the regression is lower. More specifically, the slope $s$ of the regression for any $N$ and for any $freq$ is $0 =&lt; s &lt; -( b+c)$