Balance of current through oring diodes
You are making this too complicated by not trying to understand the actual electronics.  Diodes are just one-way valves for current.  The circuit doesn't "choose" anything, as there is no high level control or logic.

Stop and actually think about what the diodes are doing.  If the power supply for one diode is higher than the output, then that diode will conduct.  Each diode does this independently.  The net result is the maximum voltage of all the supplies.

There is another wrinkle, which is that real diodes have a voltage drop when on.  The output is really the max of all the inputs, minus one diode voltage drop.

Since you are using a battery, you probably want to use the power from it efficiently.  You should use Schottky diodes, since they have a lower forward voltage drop than ordinary silicon diodes.  When the battery is at 4.2 V, for example, the output voltage might be 3.8 V.

Deliberately having a higher voltage drop in series with the 5 V supply doesn't make any sense.  When the 5 V supply is present, it seems you want the circuit to run from that instead of the battery.  Having that supply be a little higher than the max battery voltage is useful.  That insures that the current will come from the 5 V supply when available.

This does mean the circuit being powered has to be able to handle up to 5 V, as apposed to only 4.2 V.  However, that is a trivial requirement to design to.  In most cases it would take extra work to prevent it.