There is the instantaneous force between the mass and the scales, and then there's the reading of the scales. Factors influencing both of these depend on many unknown factors - so here are just some general thoughts.
First, if you drop a mass $M$ from height $h$ onto scales with mass $m$, and the two will then move as one, the collision is considered inelastic and the velocity $v'$ right after the collision can be derived from the velocity $v$ just before by putting
$$v = \sqrt{2gh}\\
v'=\frac{M}{M+m}v$$
The time that it takes to accelerate the mass is given by the elasticity of the contact, and is not in general known. The shorter the time, the larger the force - with the average force given by
$$\int F\;dt = m v'\\
F_av=\frac{mv'}{\Delta t}$$
if we consider the time-averaged force.
This force, however, probably doesn't register on your scales. I can think of three "main" kinds of scales: the balance, the spring scale, and the "return to zero" scale. None of them would register the initial impact force.
For a balance, the mass of the object is determined by comparing it to a reference mass on the other arm; that's not what you are looking for here.
For a spring scale, the deflection of a spring under the load is a measure of the weight: this is the one that is most amenable to analysis here.
For a "return to zero" scale (I can't think of the proper name - comments, anyone?) the mechanism used generates a current in an electromagnet to restore the scale to the same position it was before the mass was added. These typically have a slow response but are the most accurate - since the force is exactly proportional to the current when the scale is in the same position, there are no effects of spring ageing, temperature etc.
I believe your question can only reasonably be answered for the spring scale. Note that such scales are typically "critically damped" (although that is only approximately true since the degree of damping depends on the mass - so it can only be critically damped for one mass). But let's initially leave that out.
If we have an object with mass $M$ dropping from height $h$ onto a spring loaded scale with mass $m<<M$ and spring constant $k$, the deflection $d$ can be calculated from conservation of energy.
Total drop of the mass = $h+d$
Energy due to drop $E = Mg(h+d)$
This must be equal to the elastic energy stored at the maximum deflection:
$$\frac12 k d^2 = Mg(h+d)$$
We can rearrange and solve for $d$:
$$\frac12 k d^2 - Mgd - Mgh = 0\\
d = \frac{Mg ± \sqrt{M^2g^2 + 2 Mghk}}{k}$$
The two roots correspond to the extrema of the (simple harmonic) motion that the mass would execute: for your question, we need the positive root. Note that if $h=0$ the above shows that the maximum deflection of the scale will be twice the distance expected,
$$d = \frac{2Mg}{k}$$
and as the mass is dropped from greater height, the maximum deflection will rapidly increase more. Exactly how much more depends on the value of $k$ - the stiffer the spring, the greater the value of $k$, and the larger the value you read on the scales as the object is dropped from a greater height. All this is in agreement with your own analysis.
This is of course where the damping will come into play: a heavily damped scale will dissipate all the oscillatory energy, and will slowly reach the equilibrium value without ever overshooting. An underdamped scale will oscillate - in the limit of very light damping, it will read the value calculated above. But a critically damped scale will overshoot just once, and we can compute the amount of overshoot.
The solution for a critically damped simple harmonic oscillator is given by
$$x(t) = (A + Bt)\;e^{-\omega_0t}$$
We find A and B from the initial conditions,
$$A = x(0)\\
B = \dot{x}(0) + \omega_0 x(0)$$
If we put $x(0) = d = \frac{Mg}{k}$, the overshoot will be given by the initial velocity which we calculated above. For a light scale, this initial velocity is the velocity $v$ of the mass after falling height $h$ (I am going to ignore the additional height $d$ which I will assume to be small; otherwise the equations get a lot messier...) and the equation of motion becomes
$$x(t) = (d + (-v + \omega_0 d)t)e^{-\omega_0 t}$$
(The minus sign is there because the velocity is towards the equilibrium position). Taking the derivative with respect to $t$ we can compute the time when the overshoot is maximum and plug that back in to obtain the overshoot:
$$-\omega_0 (d + (-v + \omega_0 d)t)e^{-\omega_0 t} + (-v + \omega_0 d)e^{-\omega_0 t}=0\\
-\omega_0 (d + (-v + \omega_0 d)t)+ (-v + \omega_0 d) = 0\\
\omega_0 vt- \omega_0^2 dt + v = 0\\
t = \frac{v}{\omega_0(v-\omega_0 d)}$$
For sufficiently large $v (v>\omega_0 d)$, this will result in an extreme (overshoot).
Substituting $t$ back into the expression for $x$ will give the size of the overshoot, from which the mass registered on the balance follows.
Best Answer
In case 1, the collision between the two masses is inelastic (because they stick together), and so some of the energy is "lost", or rather, some of the energy is converted to thermal energy of the two masses. This doesn't happen in the case where you attach $m$ to $M$ when it's stationary, and hence the mechanical energy of the system is smaller in the first case then in the second case.
Quantitative analysis
In the first case, we have an inelastic collision, the result of which is that the new (maximum) speed of the oscillator is $$ v_{\textrm{new}} = \frac{M}{m+M}v\,, $$ and hence the new maximum kinetic energy (and hence total mechanical energy) is $$ K_{\textrm{max,new}} = \frac{1}{2}(M+m)\left(\frac{M}{m+M}v\right)^2 = \frac{M}{m+M} \frac{1}{2}Mv^2 <\frac{1}{2}Mv^2 = K_{\textrm{max,old}}\,. $$ In the second case, when $m$ is attached to $M$ when it is at its full amplitude, the total mechanical energy doesn't change, because the energy only depends on the amplitude, i.e., $$ U_{\textrm{max,old}} = \frac{1}{2}kA_{\textrm{old}}^2 = \frac{1}{2}kA_{\textrm{new}}^2 = U_{\textrm{max,new}}\,. $$