There is a fundamental mistake in both answers. It is not necessary to assume the autocovariance function is integrable. For a non-ergodic stationary process, it will not be integrable. The reason why Wiener gets credit for this theorem, instead of physicists like Schuster or Einstein, is that he was able to rigorously make sense of its Fourier transform anyway, in a new way, which he called «Generalised Harmonic Analysis», instead of the usual notion of the Fourier transform as given by the integral you write down. (In fact, he even anticipated Laurent Schwartz's notion of a distribution in his work on this.)
So the Wiener-Khintchine theorem states that as long as the original process $f$ is stationary and has an auto-covariance function at all, then in this new sense of Fourier transform (which works even when the Dirichlet conditions are not satisfied), the power spectral density function (which can have infinities since it is the derivative of a function which is not differentiable, and so only makes rigorous sense as a distribution) is the Fourier transform of the auto-covariance function.
==About power and finite energy signals==
If the signal has finite energy, the power is zero, as follows from your formulas below.
But only a transient signal can have finite energy. The probability of sampling a transient signal from a stationary process is zero. Transient is the exact complete opposite of stationary.
A simple unit square wave for one-cycle only has finite energy, but zero power, and as you can calculate its sample auto-covariance function easily, you see it is zero. It has to be, since the Wiener-Khintchine theorem says the Fourier transform of the auto-covariance is the power spectral density and we just saw the power is zero.
Summarizing: finite energy (which means transient) ==> zero power ==> zero sample auto-covariance function.