Magnitude characteristic of H(s)
The most direct way is to turn your transfer function into frequency response by setting $$s = j\Omega$$ and then $$H(j\Omega) = \frac{j\Omega - 1}{-\Omega^2 + 2j\Omega + 2}$$ So for $\Omega = 0$, you get $$H(0) = -\frac{1}{2}$$ which gives $|H(0)| = \frac{1}{2}$ and is in agreement with your figure. Similarly, for $\Omega = \pm 1$, you get $$H(\pm j1) = \frac{\pm j - 1}{-1 \pm 2j + 2} = \frac{\pm j -1}{1 \pm 2j}$$ You can show that $|H(\pm j1)| = \sqrt{2/5}$, which also agrees with your magnitude response plot.

However, these are only some interesting values of the magnitude response. If you need to draw the magnitude response in greater detail without using some software to do this for you, you have to go for the geometric approach. Here is a good place to start.