There are two ways to represent this, one which corresponds to an ideal model, and one which is realistic.
The idea model:
You have an ideal voltage source (with zero output impedance) with a \$50\Omega\$ resistor in series (\$R_{int}\$), and so the total output impedance is 50Ω. In your schematic above, the voltage source is ideal, and so \$R_{int}\$ would be \$50\Omega\$.
The realistic model:
The voltage source is not ideal and has some output impedance, and then the series resistor is selected so that the total output impedance is \$50\Omega\$. For instance, let's say there's an op-amp in there (quite likely). The op-amp has some output impedance, \$R_o < 50 \Omega\$, because it is not an ideal voltage source, and then the series resistor \$R_{er}\$ (internal to the function generator, not internal to the op-amp) is selected so that \$R_o + R_{ser} = 50 \Omega\$, and thus the total output impedance is \$50\Omega\$.
If you are modelling circuits in a simulator like LTSpice, you can do this two ways: you can add a \$50\Omega\$ resistor to an ideal voltage source, or you can set the series resistance of the voltage source to \$50\Omega\$. When designing a real circuit, more care is necessary and you must carefully read the datasheet to determine the output impedance of your driver.
simulate this circuit – Schematic created using CircuitLab
To sum up:
...And what does contributes to \$R_{int}\$? The function generator output impedance or the coaxial cable impedance or both?
The internal impedance of the voltage driver and a resistor inside the function generator contribute to \$R_{int}\$. The cable impedance does not contribute to this. The function generator's output impedance will be \$50\Omega\$.
