The thin-plate energy functional is defined by : $$ E = \int \kappa_1^2 + \kappa_2^2 $$ while the Willmore energy functional is defined by : $$ W = \int (\kappa_1 - \kappa_2)^2 $$
For a closed surface, if we develop $W$: $$ W = \int \kappa_1^2 + \kappa_2^2 - 2\int K = E -4\pi\chi $$ where $\chi$ is the Euler characteristic of the surface that appears following the Gauss-Bonnet theorem.
In conclusion, for a closed surface, the thin-plate and the Willmore energies differ by $4\pi\chi$.
Is it correct?
If we look at the Euler-Lagrange equations related to these energies we have : $$ \Delta^2 f = 0 $$ for the thin-plate energy, and : $$ \Delta H + 2H(H^2-K) = 0 $$ for the Willmore energy.
But since these energies differ by a topological invariant of the surface, a critical point of one energy would be also critical for the other, which means the Euler-Lagrange equations above are equivalent.
Where am I wrong ?
The biharmonic equation $\Delta^2 f = 0$ is NOT the Euler-Lagrange equation related to the energy defined by that first integral you posted.
Rather, it's what you get when you apply a thin plate approximation: $\nabla f \approx 0$. Under that, $\kappa_1^2 + \kappa_2^2 \approx f_{xx}^2 + 2 f_{xy}^2 + f_{yy}^2$. Applying Euler-Lagrange to that last expression leads to the (notably linear) biharmonic equation.