In the frequency domain, components being represented as rectangular blocks is not uncommon (resistors included!)
You already have U_a. Perhaps a small discussion concerning how the equation is produced will help.
Suppose you have a voltage divider. Would you know how to compute the output voltage (the voltage at the node between the two resistors relative to ground)? The approach is exactly the same here.
The input impedance of the op-amp is huge, so it draws negligible amounts of current. Thus, using the analysis you used for the voltage divider, you would get the same formula for the node between the two resistors in the diagram above, and since that node is connected to one of the two input terminals of the op-amp, the other input-terminal (the - terminal) has the same voltage (remember the three golden rules of op-amps!). Since the other input terminal (the - terminal) is connected to the output, we can therefore conclude that output is equal to the voltage at the node between the resistors.
Of course, the circuit above isn't the same as a voltage divider. If the input of the voltage divider is 100 V and you have two resistors of equal resistance, you get 50 V at the output. For the case of your circuit, the output voltage will be limited by the voltage you supply your op-amps with (with a 100 V input, supplying the op-amp with a measly +- 3 V will not allow the output to be 50 V!).
Concerning your comment about the resistors being 0 Ohms. The equation only holds for when at least one of your resistors isn't 0 Ohms. If you have both of the resistors at 0 ohms, then you'll short your input source! Not good.