The full analysis of this type of problem is fairly complicated and I don't have access to the figure you mention.
However, I can provide some insight. The first point is that there is a minimum threshold value, Cmin , of compression that is required to get the bottom part to budge at all. Cmin has dimensions of length.
Below, I will show that Cmin = (m / M) (M + m) g / k if the small mass part is on the top and
Cmin = (M/ m) (M + m) g/k if the large mass part is on the top. Here M and m are the masses of the heavy and light parts respectively , g = 9.81 and k is the spring constant. From this it is clear that Cm is much lower if the small mass part is on the top. Naturally, the larger is the applied compression relative to Cmin , the higher the toy will rise above the table. Further, it will be mechanically easier to apply a large C with the light mass on top without the two masses coming into contact. Taken together, it is seen the the small mass on top configuration will be the better design.
The final part of this response is a derivation of the result above: Cmin = (m/M) ( M + m) g/ k.
for the light mass on top design.
Consider a compression C, small enough such that the mass on the bottom does not move. Under this condition, the upper mass will execute simple harmonic motion (SHM) with frequency ω = sqrt( k/ m). For this SHM, the coordinate of the top part is given by x - x0 = - C cos (ω t) , where x0 is the equilibrium point about which x oscillates. It can be worked out that the coordinate y, of the center of mass is given by
y - y0 = - C (M/ (M + m) ) cos(ω t). The acceleration of the center of mass is the second derivative of this expression : aCM = C ω2 ( M/(M + m)) cos (ω t) = C (k/m) (M / (M+m)) cos(ωt ).
By Newton's second law FCM = (M +m) aCM = C k (M/m) cos(ω t)
We are most interested at the top-of-the-motion point where cos (ω t) = -1, This is the point where the downward force on the CM is its largest. The net force on the CM is FCM = N - (M + m) g, where N is the normal force exerted by the table on the system
Putting this together (for cos(ω t) = -1 and solving for N gives
N = (M + m ) g - Ck (M/m) . Since N must be greater than or equal to zero for the bottom mass not to move, this gives the condition Cmin = (m/M) (M + m) g/k.
The same analytic approach gives Cmin = (M/m) (M + m) g/k for the heavy mass on top design.