Although it may appear that the mean of the log-transformed variables is preferable (since this is how log-normal is typically parameterised), from a practical point of view, the log of the mean is typically much more useful.
This is particularly true when your model is not exactly correct, and to quote George Box: "All models are wrong, some are useful"
Suppose some quantity is log normally distributed, blood pressure say (I'm not a medic!), and we have two populations, men and women. One might hypothesise that the average blood pressure is higher in women than in men. This exactly corresponds to asking whether log of average blood pressure is higher in women than in men. It is not the same as asking whether the average of log blood pressure is higher in women that man.
Don't get confused by the text book parameterisation of a distribution - it doesn't have any "real" meaning. The log-normal distribution is parameterised by the mean of the log ($\mu_{\ln}$) because of mathematical convenience, but equally we could choose to parameterise it by its actual mean and variance
$\mu = e^{\mu_{\ln} + \sigma_{\ln}^2/2}$
$\sigma^2 = (e^{\sigma^2_{\ln}} -1)e^{2 \mu_{\ln} + \sigma_{\ln}^2}$
Obviously, doing so makes the algebra horribly complicated, but it still works and means the same thing.
Looking at the above formula, we can see an important difference between transforming the variables and transforming the mean. The log of the mean, $\ln(\mu)$, increases as $\sigma^2_{\ln}$ increases, while the mean of the log, $\mu_{\ln}$ doesn't.
This means that women could, on average, have higher blood pressure that men, even though the mean paramater of the log normal distribution ($\mu_{\ln}$) is the same, simply because the variance parameter is larger. This fact would get missed by a test that used log(Blood Pressure).
So far, we have assumed that blood pressure genuinly is log-normal. If the true distributions are not quite log normal, then transforming the data will (typically) make things even worse than above - since we won't quite know what our "mean" parameter actually means. I.e. we won't know those two equations for mean and variance I gave above are correct. Using those to transform back and forth will then introduce additional errors.
Best Answer
The Wikipedia statement
is wrong.
OLS does NOT have assumptions on response variable. But has assumptions on residual (See Gauss–Markov theorem). Also see this post for details.
Why linear regression has assumption on residual but generalized linear model has assumptions on response?
I am stealing @Cliff AB 's example here. The following distribution on $y$ and residual does not violate OLS assumption!
Related posts:
What is a complete list of the usual assumptions for linear regression?
How does linear regression use the normal distribution?
What if residuals are normally distributed, but y is not?