The usual exercise is to get the student to work out the distribution of sample variances.
But I think the confusion arises over the terms used vis: the "sample variance" is a technical terms that does not quite mean the same thing as "the variance of the sample", but the "population variance" is the same thing as "the variance of the population".
The sample variance is an approximation to the population variance which is agreed upon by convention. The division by (n-1) gives a better approximation than the division by n (which would have given you the variance of the sample).
To see what they are doing: remember that the idea is to figure out what the population mean and variance is without actually polling the entire population. You could take a sample of 1000 out of a population of several million ... what can you say, in general, about the entire population, from such a small number?
You could find the mean and variance for the sample ... OK. But if you took another sample of 1000 tomorrow you will very likely get a different mean and variance from them.
If you take a lot of samples, and they are all random, then you get the mean-value-theorum giving you a distribution of means and variances which are, themselves, normal distributions.
If the population was normally distributed, then the mean of the means will get closer to the population mean as the number of samples increases but the mean of the variances (of the sample) will be bigger than the population variance.
You should be able to confirm that by working them out for just three or four random normal variables. You should know how to add random distributions by now.
What the passage quoted is saying is that if you define the sample variance to divide by (n-1) it is more convenient for estimating the population variance which is what we are after.
We don't have to do it that way, it's a convention.