Bridge rectifier circuit
I don't get what is the relation between the dc output voltage and Vs


Firstly, the output voltage is not "+,- 15V" - it is +15V dc (relative to ground/earth). 

Secondly, the "+" and "-" labels on the transformer secondary winding are meaningless in this context - the secondary voltage is alternating and labeling them as "+" and "-" is confusing.

If your output dc voltage is +15V then, in simple terms, the secondary ac voltage has a peak value of about 15V. If your secondary voltage was less then your dc output voltage would be less.

Diode volt-drop - diodes aren't perfect conductors of electricity in one direction; they do "lose" about 0.7V and this means for a +15V dc output, the ac voltage feeding the bridge is more like 16.4V peak.

To calculate what the transformer secondary ac RMS voltage is, you divide the calculated peak voltage (16.4V) by \$\sqrt{2}\$ and this is 11.6V RMS.

Ripple - if you didn't have a load resistor, the capacitor would become charged to 15V and remain at that voltage. However, the resistor discharges the capacitor over every half cycle of the AC secondary voltage - this is what causes ripple voltage: -



To calculate ripple you need to know what R and C values are and here's another picture with a more detailed analysis: -