I'm no expert, but perhaps I can give you a start.
I think the confusion is between the bandwidth (BW) of the signal and the BW of the filters.
When a radio signal is created, a nominal carrier frequency is modulated with an information
signal and new frequencies called sidebands are created. There are several ways of doing this: a
radio wave can have it's amplitude, frequecy and phase varied, or a combination of them.
Probably the simplest example is amplitude modulation (AM). Say an audio signal containing
frequencies from 300Hz to 12kHz is modulated onto a carrier at 1MHz. A 12kHz tone will generate
frequencies at 0.988MHz and 1.012MHz. A 300Hz tone will generate frequencies at 0.997MHz and
1.003MHz. Other audio tones will generate other frequencies in this range 0.998-1.012MHz, meaning
that the radio signal will be composed of a mixture of frequencies in this range - a bandwidth of
24kHz centred on 1MHz.
For the reciever to reconstruct the original audio signal, it need to receive all these sideband
frequencies and to demodulate them.
If the receiver has filters which are narrower than the BW of the signal, some sideband
frequencies are lost and the reconstructed signal is distorted. That may or may not matter,
depending on the context and on the degree of distortion. Some distortion is inevitable.
If the reciever has filters which are wider than the BW of the signal, then extra noise and other
unwanted signals are allowed through, which again may reduce the quality of the final demodulated
signal.
Generally the results will be best when the BW of the filter is matched to the BW of the signal to
get the maximum of wanted frequencies and the minimum of unwanted frequencies. Sometimes there is
a bit of tradeoff, where reducing the BW of the filter looses a small amount of desired signal but
a lot of undesired interference. The result is not pefect, but optimum.
Now why are the filters in your receiver not all the same BW? Roughly one could say that, for a
given amount of cost, effort or complexity, the BW of a filter is proportional to it's centre
frequency. A filter with a 2% BW would be 30kHz wide at 1.5MHz, 300kHz wide at 15MHz and 40MHz
wide at 2GHz. If you need to have a filter 30kHz wide, this will be easier to do at the lower
10.7Mhz frequency - one reason for converting frequencies before demodulation. (The reason for
10.7MHz is simply that this is a standard frequency for which one can buy readymade filters.)
The receiver BW is that of the narrowest filter, usually at the lowest IF. Filters are desirable at the earlier stages, mainly to reduce intermodulation effects and image frequencies, but they usually don't need to be very narrow and it would not be worth the cost to make them as narrow as the last filter. And they may need to be much wider than the BW of any one signal.
If a receiver needs to receive a range of frequencies (eg. your FM radio needs to receive signals from about 88MHz to 108MHz) then the RF filter and any IF filters before the mixer with the variable frequency input, need to accomodate this range of frequencies. In your case, I guess the 1930-1990MHz bandwidth represents a range of possible signals each of which is only 30kHz wide. Your first stage needs to accept any of these signals, then the 2nd mixer chooses one of them by mixing with a particular frequency to step down to 10.7MHz. The 10.7MHz filter then allows just this one signal with it's 30kHz of sidebands to pass through to the demodulator. The frequencies shown in your diagram select the signal at exactly 1960MHz and it's sidebands from 1959.970MHz - 1960.030MHz. If you changed the 2nd local oscillator frequency to, say, 933.05MHz then you'd receive a signal on 1943.75MHz and it's 30kHz of sidebands. Both of these signals (as well as hundreds of others) would pass through the earlier 60MHz BW filters.
To try to clarify how the signal is being reduced in frequency, I'll work through it. The basic point to have in mind is that, to a first approximation, a mixer takes two input frequencies and produces two outputs at the sum and the difference of the frequencies. Say I mix 40MHz and 50MHz, then I get difference=10MHz and sum=90MHz out.
In your case, 1960MHz is mixed with 1000MHz producing difference=960MHz and sum=2960MHz. Your first IF filter is centred at 960MHz, so the 60MHz BW allows 930MHz-990MHz through. The 2960MHz is way too high and is lost. Your second mixer combines this 960MHz with 949.3MHz to produce difference= 10.7MHz and sum= 1909.3MHz. The 2nd IF filter based on 10.7MHZ with 30kHz BW allows 10.685-10.715MHz to pass through. Again, 1909.3MHz is way too high and is lost, and the 10.7MHz signal with it's sidebands gets through.
If, as I said above, we changed the 2nd local oscillator to 933.05MHz, then the first mixer stays the same, but the 2nd mixer now combines 960MHz with 933.05MHz producing difference= 26.95MHz and sum=1893.05MHz. Neither of these frequencies can get through the 10.7MHz filter and this signal is blocked. But another signal which was transmitted on 1943.75MHz would go as follows:
mix with 1000MHz gives sum=2943.75MHz and diff=943.75MHz. The latter would get through the 1st IF filter (930-990MHz). So the 2nd mixer gets 943.75MHz and 933.05MHz to give sum=1876.8MHz and diff=10.7MHz. Again the later gets through the 2nd IF filter and is demodulated to give the signal that was transmitted on the 1943.75 carrier.
So by varying the 2nd local oscillator you can select which signal you receive from throughout the 1930-1990MHz band.
