The commonly referred-to as natural variables of a thermodynamic potential are those which appear "naturally" in its total differential. For example, from the first principle of thermodynamics, the internal energy satisfies:
$$
dU=TdS-pdV+\mu dN
$$
thus $U=U(S, V, N)$ is expressed in such a way that any thermodynamic function of the system is expressed as derivative of the internal energy:
$$
T=\left(\frac{\partial U}{\partial S}\right)_{V,N}\qquad
p=-\left(\frac{\partial U}{\partial V}\right)_{S,N}\qquad
\mu=\left(\frac{\partial U}{\partial N}\right)_{V,S}.
$$
Upon action of the Legendre transform $H=U+pV$ enthalpy has:$$
dH=TdS+Vdp+\mu dN \implies H=H(S,p,N).
$$
And so on: Helmholtz free energy $F(T,V,N)=U-TS$, Gibbs free energy $\Phi(T,p,N)=H-TS$ and the grandpotential $\Omega(T,V,\mu)=F-\mu N.$
The use of such functions is determined by the experimental setting you can produce for a given system: for example a low-density gas can be easily held at a given volume and given number of particles, at a certain temperature, for it can be enclosed in a box and its volume varied with a piston. Thus it is natural to use Helmholtz free energy, which naturally embeds $T,V,N$ as free variables.
Other materials might be less easy to compress or dilate, so one might choose to vary their pressure, which would lead to using Gibbs free energy in the variables $T,p,N$.
The example you quoted is correct exactly in this sense: if you know the functional dependence of $U$ on its natural variables $U=U(S,V,N)$ then you can use the formulas listed above; but if you have, say, $U=\tilde{U}(T,V,N)$ then how do you calculate the equilibrium entropy $S$?
You have to switch to another thermodynamic potential: $F=F(T,V,N)$ using the Legendre transform:
$$
F=F(T,V,N)=U\big(S(T,V,N),V,N\big)-TS(T,V,N)
$$
where $S(T,V,N)$ indicates the expression of $S$ obtained by inversion of the relation
$$
T=\left(\frac{\partial U}{\partial S}\right)_{V,N}.
$$
