The two forms of standard deviation are relevant to two different types of variability. One is the variability of values within a set of numbers and one is an estimate of the variability of a population from which a sample of numbers has been drawn.

The population standard deviation is relevant where the numbers that you have in hand are the entire population, and the sample standard deviation is relevant where the numbers are a sample of a much larger population.

For any given set of numbers the sample standard deviation is larger than the population standard deviation because there is extra uncertainty involved: the uncertainty that results from sampling. See this for a bit more information: Intuitive explanation for dividing by when calculating standard deviation?

For an example, the population standard deviation of 1,2,3,4,5 is about 1.41 and the sample standard deviation is about 1.58.

There are, in fact, two different formulas for standard deviation here: The population standard deviation $\sigma$ and the sample standard deviation $s$.

If $x_1, x_2, \ldots, x_N$ denote all $N$ values from a population, then the (population) standard deviation is $$\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2},$$ where $\mu$ is the mean of the population.

If $x_1, x_2, \ldots, x_N$ denote $N$ values from a sample, however, then the (sample) standard deviation is $$s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \bar{x})^2},$$ where $\bar{x}$ is the mean of the sample.

The reason for the change in formula with the sample is this: When you're calculating $s$ you are normally using $s^2$ (the sample variance) to estimate $\sigma^2$ (the population variance). The problem, though, is that if you don't know $\sigma$ you generally don't know the population mean $\mu$, either, and so you have to use $\bar{x}$ in the place in the formula where you normally would use $\mu$. Doing so introduces a slight bias into the calculation: Since $\bar{x}$ is calculated from the sample, the values of $x_i$ are on average closer to $\bar{x}$ than they would be to $\mu$, and so the sum of squares $\sum_{i=1}^N (x_i - \bar{x})^2$ turns out to be smaller on average than $\sum_{i=1}^N (x_i - \mu)^2$. It just so happens that that bias can be corrected by dividing by $N-1$ instead of $N$. (Proving this is a standard exercise in an advanced undergraduate or beginning graduate course in statistical theory.) The technical term here is that $s^2$ (because of the division by $N-1$) is an unbiased estimator of $\sigma^2$.

Another way to think about it is that with a sample you have $N$ independent pieces of information. However, since $\bar{x}$ is the average of those $N$ pieces, if you know $x_1 - \bar{x}, x_2 - \bar{x}, \ldots, x_{N-1} - \bar{x}$, you can figure out what $x_N - \bar{x}$ is. So when you're squaring and adding up the residuals $x_i - \bar{x}$, there are really only $N-1$ independent pieces of information there. So in that sense perhaps dividing by $N-1$ rather than $N$ makes sense. The technical term here is that there are $N-1$ degrees of freedom in the residuals $x_i - \bar{x}$.

For more information, see Wikipedia's article on the sample standard deviation.