There is no simple answer.. Take a computer system with a 500W PSU. The sum of all parts may be capable of dissipating 1000W but utilization due to usage may only be 100W on the average. The best way is to plan for more capacity and then cost reduce later after testing it.
It is much more complicated than your suggestion. Current drawn depends on voltage, temperature . For sure you can estimate the nominal load current ought to be less than worst case load current. Unless you know something about the load curent vs voltage and determine what the Power Fail threshold when your system should shut down safely, otherwise you can let it stop in unreliable ways.
So you have no favourable measure of accuracy of predicting the operating time without experience or schematic and description to someone who understands.
We also have no idea what your design is and if you have any power fail circuit detection like the low voltage detection built into MOBO's
Sorry without more there is no simple answer to a software expert. It would be like a hardware guy telling howe many lines of unique code gets executed per minute. You could look at all the code and features and multiply by some magic utilization ratios. But the chance of it being accurate is inversely proportion to its complexity... Most like all you have to deal with is the power consumption of the top 3 items like the motors. But the the backup time is limited by the weakest link and we do not know if your power backup is balanced for each subsystem of batteries.
If the programmer does not supply power (OFF): should I connect the +5V and GND pins from the programmer to the same +5V and GND as the target board, which is powered by another source?
The jumper most likely physically disconnects the +5V pin, so it doesn't matter. Keep GND connected in any case, because the devices require a common reference level to guarantee a reliable communication.
If the programmer does supply power (ON): should I power the target board completely from the programmer, so that I pull the +5V and GND on the connector to VCC and GND on the target board? If so, the programmer would be the only power supply.
Yes, its probably best to completely disconnect the external power supply in that case. Connecting two different DC voltage sources in parallel is almost always a bad idea. Keep in mind, that USB (2.0) can only supply 500mA@5V max, while the programmer will also consume a few mA itself.
Best Answer
Although the article you linked to doesn't mention it, other articles like this one do confirm that the device gets its power using RFID technology. That is the way a passive RFID employee ID badge, for example, can work without needing a battery installed.
Besides being used to receive and transmit the RF frequencies used for communication with the tiny transceiver IC, the antenna is also used to receive a relatively higher-powered RF signal that is rectified and used to power the circuit (this article mentions a capacitor). I have no idea what the power requirement is for the Google contact lens device, but a typical power level for a passive RFID tag at maximum range is 1 µW -- 1 V at 1 µA.