There must be at least $ 5 $, and this is easy to see if you picture the graph of $ g $. The graph of $ g $ is symmetric across the line $ x = 1/2 $ and has a local maximum at $ x = 1/2 $. We are given that there is a global maximum elsewhere, so there must actually be at least two global maxima, since if a global maximum occurs at $ x = a $, another one must occur at $ x = 1 - a $ by symmetry. Moreover, since $ x = 1/2 $ is a local maximum, it follows that between $ x = 1/2 $ and the two global maxima, there must be two local minima as well. Therefore, this gives us at least $ 5 $ points on which the derivative of $ g $ vanishes. (You can make this argument precise using continuity and Rolle's theorem.)
To see that this bound is tight, note that $ f(x) = (x+1)(x+1/2)(x+2/3) $ satisfies the desired conditions, and $ g'(x) $ is a polynomial of degree five, so it can have no more than five roots; and the five roots come from the two local minima and the three local maxima of $ f(x - x^2) = g(x) $.