How do we choose the variables which the state function's depend on?

Let's start with internal energy. Internal energy is "naturally" a function of the three extensive variables $S$, $V$, and $N$. So it's actually not $U(V,T)$, instead it's $U(S,V,N)$. The reason this is "natural" is because the intensive variables $T$, $P$, and $mu$ are given directly by derivatives of the internal energy, with the other two variables held constant:

$$T=left(frac{partial U}{partial S}right)_{V,N}$$ $$-P=left(frac{partial U}{partial V}right)_{S,N}$$ $$mu=left(frac{partial U}{partial N}right)_{S,V}$$

So every one of the thermodynamic variables is either an independent variable of $U$, or the first derivative of $U$ with respect to an independent variable. We call functions with this property thermodynamic potentials.

Thermodynamic potentials are particularly nice quantities to define because, if we have control over the three explicit variables, the values of all thermodynamic variables can be directly obtained just by varying those three variables and measuring the change in that potential. So, if you have control over $S,V,N$, and you can measure $U$, then all you need to do is change the appropriate one of $S,V,N$ a bit while holding the other two constant, and read the slope of $U$ in each direction to get your intensive variables.

If instead we suppose that $U$ is an explicit function of $T,V,N$, then we don't get the other three variables back when we take the first derivative:

$$left(frac{partial U}{partial T}right)_{V,N}=C_V$$ $$left(frac{partial U}{partial V}right)_{T,N}=Tleft(frac{partial P}{partial T}right)_{V,N}-P$$

So $U(T,V,N)$ is not a thermodynamic potential. It is a state function which can be used in conjunction with the thermodynamic potentials to constrain the properties of a system, but it is not the "natural expression" of internal energy, and will not allow you to relate the internal energy to other thermodynamic potentials.

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Now that we have defined the "natural expression" of internal energy as $U(S,V,N)$, then we can derive the other thermodynamic potentials as Legendre transformations of this one. Applying these transformations, we get:

$$F(T,V,N)=U-TS$$

$$H(S,P,N)=U+PV$$

$$G(T,P,N)=U-TS+PV$$

(Note: this is not an exhaustive list. There are actually 7 possible transformations if you also start transforming $N$ into $mu$, but those aren't as commonly used.)

It should be fairly easy to check that the above property also holds for these variables: for example, for the Gibbs free energy $G$, we have that

$$-S=left(frac{partial G}{partial T}right)_{P,N}$$

$$V=left(frac{partial G}{partial P}right)_{T,N}$$

$$mu=left(frac{partial G}{partial N}right)_{T,P}$$

And, like before, if you have control over $T,P,N$ and can measure $G$, then you can easily "read off" the values of $S,V,mu$ by tracking the change in $G$ due to small perturbations in $T,P,N$.