Clarification on how bandwidth effects input signal
The system $H(f)$ appears to be a lowpass filter. As you state, the fact that the bandwidth decreases results in a more distorted output. However, I would not say that "fewer segments of the signal pass through the system".

The Fourier transform of the input in your image is a $\mathrm{sinc}()$ function. This function extends to $\pm \infty$, so to get absolutely no distortion you would need an infinite bandwidth. As this is not possible, then distortion will always appear. Depending on where we "cut" the sinc's "tails", the output will be more or less distorted. If we let "more frequencies" pass, then the output will look more like the input.

The square pulse has very sharp edges - it suddenly goes from $0$ to $1$. This speed translates to high frequency components. As $H(f)$ is a lowpass filter, high components will be filtered out. To what extent this will affect the signal will depend on the bandwidth. As it decreases, more high frequencies will be cut off, leading to a slower transition at the output, which leads to that smoothed version of the square pulse.