What are the differences between sd of population divided by sqrt of sample size and sd of sample divided by sqrt of sample size? What are the differences between the standard deviation of a population divided by the square root of the sample size and the standard deviation of a sample divided by the square root of the sample size? Are they both called standard error?

Yes, they are both measures of standard error for the population average, i.e. they can be considered as measures of standard deviation for the sampling distribution of the sample average.
More precisely, consider an independent random sample $X_1,\ldots, X_n$ with $E(X_i) = \mu$ and $\text{var}(X_i) = \sigma^2$, both assumed to be finite. Then the sample average $\bar X = n^{-1}\sum_i X_i$ has expectation
$$E(\bar X) = \mu$$
and variance
$$
\text{var}(\bar X) = \frac{\sigma^2}{n}.
$$
Thus $\sigma/\sqrt{n}$ is really the standard deviation of $\bar X$. Assuming the sample is large enough, we can invoke Central Limit Theorem to obtain the asymptotic pivot
\begin{align*}
\frac{\sqrt{n}(\bar{X}-\mu)}{\sigma^2}\, \,\dot\sim\,\, N(0, 1), \tag{*}
\end{align*}
where $\dot\sim$ is to be read approximately distributed as. If the population variance is known, we can use (*) to build a confidence interval for $\mu$
of approximate level $1-\alpha$
$$\bar X \pm z_{1-\alpha/2}\sqrt{\frac{\sigma^2}{n}}.$$
However, such an interval is not usable if $\sigma^2$ is unknown. But, since $S^2 = (n-1)^{-1} \sum_i (X_i-\bar X)^2$ is an unbiased estimator for $\sigma^2$, then we can think of replacing the unknown $\sigma^2$ by its unbiased estimate $S^2$, giving thus the alternative interval
$$
\bar X \pm z_{1-\alpha/2}\sqrt{\frac{S^2}{n}}.
$$
Thus, $\frac{S^2}{n}$ is a replacemnt of $\frac{\sigma^2}{n}$ when the population variance is unknown.
