The Lagrangian of the system is, $$L = \frac{1}{2}m\dot{x_1}^2 - \frac{1}{2}k_1(l_1 - l_0+x_1)^2-\frac{1}{2}k_2(l_2-l_0+x_2-x_1)^2+mgx_1$$
Here, $x_1$ is the downward distance from the equilibrium position of the mass, $x_2$ is the downward distance from the midpoint of the driving oscillator, $k_1$ and $k_2$ are spring constants of the top and bottom springs respectively, $l_0$ is the unstretched length of both springs, $l_1$ and $l_2$ are stretched lengths at equilibrium of the top and bottom springs respectively. The condition for equilibrium, when the driving force is not applied, is $$k_1(l_1-l_0) = k_2(l_2-l_0)+mg$$ We can take $x_2=a\ cos(\omega t)$ for the driving oscillation and then the equation of motion for the mass from Euler-Lagrange equation would be, $$m\ddot{x_1}+(k_1+k_2)x_1-ak_2cos(\omega t) = 0$$ Solving this equation, we can get the amplitude of the driven oscillation to be, $$A=\frac{ak_2}{k_1+k_2-m\omega^2}$$ As a result, the resonant frequency should be the same for both setups with springs interchanged as $\omega_r=\sqrt{(k_1+k_2)/m}$. The proportion between the resonant frequencies in your setups should be $\omega_{r(5,5)}:\omega_{r(5,15)}:\omega_{r(15,15)}=1:\sqrt{2}:\sqrt{3}$ which is apparent in your results. However, based on theoretical calculations the resonant frequencies should be the same for setups with just the springs interchanged.
Now, in this calculation damping is ignored while in all real systems damping is inevitable. The presence of damping is also the reason you are not getting an infinite amplitude at resonance. For a damping factor $\gamma$, the resonant frequency will be $$\omega_r = \sqrt{\frac{k_1+k_2}{m} - \frac{\gamma^2}{2}}$$
If the damping is not same for both the springs, then the resonant frequencies might not match within a certain limit. I would suggest you to check for any possible source of damping in this setup that might significantly affect the resonant frequency.