It's easy to overthink this. For example - When applying the Left TS switch, (+) hits the Left TS bulbs AND one side of the indicator bulb. This is a parallel circuit. The other side of the Left TS bulbs is solidly bolted to battery negative (-).
The Left TS bulbs send the (+) to ground (-) thru their filaments, there is 12V dropped across each TS filament, max operating current, they light up.
The Indicator filament is also connected to the Right TS bulbs which are not energized(+) when the left turn is selected. The other side of the Right TS bulbs is solidly bolted to battery negative (-).
The (+) applied to the TS Indicator passes thru it, then thru the Right TS bulb filaments and then goes to ground (-); this is a series circuit. 12V drops across the combined filaments on its way back to (-) Ground.
The Indicator bulb has a high filament resistance compared to the turn signal bulbs. This results in the majority of the voltage dropping across the indicator filament. The Indicator bulb resistance also largely determines this portion of the circuit current flow. So the Right TS filaments don't have enough current to glow, but the Indicator does.
LED lamps are complicated devices. A LED has to use resistance to control current magnitude. Depending on internal construction, they can be directional or not. A single Regular diode is used to make them directionally protected, they'll only glow if wired in properly. BUT - a "diode bridge" will allow the LED to operate "bi-directionally", the bridge automagically reroutes the applied DC (or AC) voltage to always be the proper polarity across the LED. All of this hardware can be built into an LED "lamp". SO - IF the LED bulb is a "Bi-Directional" lamp, it'll work fine in your TS Indicator.
As far as a "4-Way Flasher", jumping either the Indicator or the G/W & G/B with a small toggle switch would get you there.