Modal analysis and frequency response function - interpretation needed
This may be a better question for https://physics.stackexchange.com/ but I'll give it a shot.


I assume you show the FFT of the time waveform at the sensor. In order to get the actual frequency response you would need to divide by the FFT of the hammer response. Since the hammer is reasonably smooth, we can eyeball this. It also looks like your measurement is "good": sufficient signal to noise ratio and nothing unexpected in there.
Force is a vector: it's unclear whether you are only measuring the normal force or a full 3 component vector and calculate the magnitude. 
At a broad brush, the frequency response can be interpreted as "reasonably flat" with a bunch of local wiggles at specific frequencies
The flat part of the frequency response (FR) simply means the piece itself is moving (or compressing) uniformly: i.e. it's moving as one without the sensor surface deforming.
The wiggles are modal behavior. Here the piece (or surface) has different movement at different locations. 
The shape of the modes is determined by the geometry (and material) of the piece. That's why the modes tend to be very narrow band.
Not all modes will be visible in a measurement like this. Modes have "nodes" and "peaks" as a function of position. In order to see mode both excitation and sensor need to be far enough away from a node. If the hammer is close to the node, the mode will not be excited and if the sensor is close to the node, the sensor won't pick it up (despite it being there and active).
Modal movement implies different areas of "positive" and "negative" movements. 
The modal movement simply adds to the "move as one piece" movement. If both go in the same direction you see a peak and if both go in opposite directions you see a dip.



  Does it mean that forces measured at these frequencies are damped?


Not really: what it means is that movement due to modal behavior and movement of the entire piece as a single chunk are in opposite directions at this particular point and frequency. So the sum is smaller than each individual movement.