Why do you subtract the mean when calculating autocorrelation? Autocorrelation is defined as this:
$$\widehat{\rho}_{k} = \frac{\sum_{t=k+1}^{T}{\left(r_{t}-\overline{r}\right)\left(r_{t-k} - \overline{r}\right)}}{\left(\sum_{t=1}^{T}{r_{t} - \overline{r}}\right)^{2}}$$
Why is it necessary to subtract from the mean?
Is it not sufficient to divide the value at $k+1$ by the value at $k$ instead?
$$\rho = \text{mean}\left(\frac{r_{k + 1}}{r_{k}}\right)$$

Let's start from the basics. Variance tells us about the variability around the mean
$$
\operatorname{Var}(X) = E[(X - E[X])^2]
$$
You can generalize this concept to two variables, the covariance
$$
\operatorname{Cov}(X, Y) = E[(X - E[X]) (Y - E[Y])]
$$
where variance is a special case of it
$$
\operatorname{Cov}(X, X) = E[(X - E[X])^2]
$$
Correlation is just a normalized covariance so that it is bounded between -1 and 1,
$$
\operatorname{Corr}(X, Y) = \frac{\operatorname{Cov}(X, Y)}{\sigma_X \sigma_Y}
$$
Autocorrelation is just a special case of correlation.
Yes, you can calculate the expected value of the ratio of two variables and in some case, it might be a meaningful statistic, but it doesn't anymore measure the "spread" or "co-spread" of the variables.
You may be interested in reading the How would you explain covariance to someone who understands only the mean? thread.
