Transfer characteristic of two-cascade resistive voltage divider
If you had a buffer amplifier with gain of 1 between the stages then the first equation would be correct (assuming no load on the output).
Since the second voltage divider loads the first one, the output signal is affected, and obviously will be less than in the non-interacting case.
The loaded output of the first voltage divider becomes Rx/(Rx + R1) where Rx = R2 || (R3+R4).
Remaining algebra is left as an exercise.

Edit: As mentioned above, the loading of the first network by the second alters the transfer function so we cannot simply modify the individual (unloaded) G1(s), G2(s) transfer functions to get the result G1(s)*G2(s) for the total transfer function.