Should each trace carrying RF be 50Ohm in characteristic impedance? How?
I only have a vague idea of what characteristic impedance


Characteristic impedance is the ratio of voltage to current (thus, an impedance) for signals propagating along the trace, which is determined by the balance of capacitance and inductance along the trace.


  It should be dependent on the length and the frequency, how come it isn't?


Characteristic impedance depends on the ratio of inductance to capacitance. Since both inductance and capacitance increase linearly when the trace length increases, their ratio doesn't depend on the trace length. 

Also, within limits, these parameters also don't change much with frequency, so again the ratio doesn't depend on frequency and the characteristic impedance doesn't depend on frequency. 


  Intuitively I should calculate the characteristic impedance of each pad-to-pad trace and make sure it is always 50Ohm. Is that the case?


If the driving circuits are designed to drive 50 ohm loads, then generally yes. You also want to provide matched termination at at least one end of the trace, and possibly both, depending on the details of your circuit.

Generally you don't have to make a separate calculation for each connection. You just look at your board stack-up and find a trace width that achieves 50-ohm characteristic impedance, and make all of your traces that width. You might use microstrip, stripline, or coplanar waveguide geometry depending on the circumstances of your layout. You would do a separate calculation for each signal layer on your PCB, and maybe for the different types of geometry (microstrip and coplanar, single-ended and differential) if you need to use all those combinations.

If the trace length is less than about 1/10 of a wavelength at your operating frequency, then you can often get away with using an unmatched trace.