I think you (Marmitus) should maybe have a look at webpages concerned with MOSFET/FET spec and use. Loads of people misunderstand it so you're not alone
I assume you think Vgs = Voltage Gate Saturation?
No.
There is no saturation like BJTs (bipolars) just as said by Premelec.
The 'threshold' voltage is an agreed spec which means the Gate Voltage (wrt Source) required to permit a current of 250uA to flow. You'll see this on all MOSFET transistor Data Sheets. It does NOT mean that is the maximum permitted voltage at the Gate. And it does NOT mean the threshold at which the MOSFET acts like a 'perfect' switch.
In fact, it is clearly shown in the Data Sheet that a Vgs of 4.5V results in an Rds(on) of 2 to 5.3 ohms (in the ST Data Sheet).
The absolute MAximum Gate-Source voltage is +/- 18V as shown in the Data Sheet.
So, you don't need a pot/div. And as you haven't got one in your circuit that's fine.
To determine the Gate voltage required you should look at the graphs. Example "Output Characteristics" in the Data Sheet.
They say Gate-Source Voltage (Vgs) as you should have the MOSFET source connected to Ground for a simple switch like you want. Therefore everything is referenced to Ground (0V).
I would assume you will be switching tiny currents so your circuit will be fine, though I don't know what happens
after your connector.... so it may not.... remember your circuit will act as a low-sided switch (switch to ground or switch-to-lower-voltage-line).
For greater consistency generally you can get MOSFETs which are better for switching at 5V - Logic Level.
A MOSFET is not a general purpose relay substitute. When used properly it can replace a relay but you can't connect power any-old-way-round.