I have looked at several derivations online, and the clearest I have found is that on the Wiki page.
The problem being considered is that of two phases, with an interface of a given area between them. The Gibbs adsorption isotherm relates the change in the surface tension to the change in chemical potential in equilibrium, via the excess surface concentrations.
The result for a two species system,
$-d\gamma = \Gamma_1 \mu_1 + \Gamma_2 \mu_2$,
where $\mu_i$ are the chemical potentials of the species at the surface, and thus also in each bulk phase in equilbirum. $\Gamma_i$ is an 'appropriately defined surface excess', the significance of which I don't quite get, but I see that it measures a number density giving the per unit area of the species.
A more familiar form might be
$\Gamma_i = -\frac{1}{RT}(\frac{\partial \gamma}{\partial \ln(C_i)})_{T,p}$.
The part of the derivation which I don't understand is right at the start, where the Gibbs free energy for a given phase is stated to be
$G=U-TS+PV+\Sigma_i \mu _i n_i.$
I thought that $U=TS-PV+\Sigma_i \mu_i n_i +\gamma A$, so with the above definition of $G$ we have an extra $\Sigma_i \mu _i n_i$ here! However the derivation does not seem to work without it.
Another similar problem is that the $\gamma A$ term seems to be given additionally to the Gibbs free energy, aside from that implicitly contained in $U$. Again, I am not quite sure why we are putting in these extra pieces.