Impedance of BNC vs output impedance of function generator
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 &lt; 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 &ndash; 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\$.