The image from the data sheet is a little hard to read, but you will apply a voltage of up to 30 volts between +Ub and -Ub. Then you will see a voltage between S+ and S- proportional to the pressure.
3.5 mV/V is the output of the sensor for its full range (maximum value). So if you were to apply your 3.3 volts to the sensor, your maximum signal before amplification would be 3.3 volts * 3.5 mV or 11.55 mv for 10 bar.
For example, a one-bar input (10% full scale) would give an output of 1.115 mv. The two output pins are each going to be near 1/2 the supply voltage, or 1.65 volts. It will appear to be 1.651115 on S+, and 1.648885 on S- to give you your 1.115 millivolt differential voltage.
Since the each sense element's resistance is 10K, the output impedance is 5K on each of these signals, so the signals must be amplified with a high-impedance, high gain instrumentation amplifier before reading by the ADC.
The image comes from a TI reference design for an INA333 which is one I use. It has a good discussion of how to get an accurate reading from a strain gauge like the one you are using. See https://www.ti.com/lit/ug/tidub00/tidub00.pdf?ts=1682202573400