I'm answering my own question but I think that I have an intuitive justification for why the sample standard deviation reflects the population standard deviation.
As our sample size becomes larger and larger, it better reflects the population, and thus the variation of the population. I realize that we could say the same for the sample mean, but at least for the case of flipping coins, the sample s.d. seems to tend adhere more towards the pop s.d. compared to sample mean.
Suppose we have a true 50 50 coin: p=0.5 q=0.5. Half of all flips in the coin's history (q = 0.5) give tails = 0, and half of all flips in the coin's history (p = 0.5) give heads = 1. The population mean of all flips in coin's history = 0.5, and Standard Deviation = root (0.5*0.5) = 0.5.
if flip 10 times and get q=0.6, p=0.4 for 1 sample
then sample mean = 0.4, SD = root (0.6 x 0.4) = 0.490
sample mean has decreased 20% from pop mean, but SD decreased 2%
if flip 10 times and get q=0.7, p=0.3 for 1 sample
then sample mean = 0.3, SD = root (0.7 x 0.3) = 0.458
sample mean has decreased 40% from pop mean, but SD decreased 10%
if flip 10 times and get q=0.8, p=0.2 for 1 sample
then sample mean = 0.2, SD = root (0.8 x 0.2) = 0.4
sample mean has decreased 60% from pop mean, but SD decreased 20%
suppose flip 10 times and get q=0.9 p = 0.1 for 1 sample
then sample mean = 0.1, SD = root (0.9 * 0.1) = 0.3
sample mean has decreased 80% from pop mean, but SD decreased 40%
This scenario shows that over the vast majority of the sampling distribution of possible sample means of 10 flip outcomes, the sample standard deviation does not stray too far from 0.5, even though the sample mean can vary much more
Suppose you want to estimate the mean $\mu$ of a normal population using the
mean $\bar X$ of a random sample $X_1, X_2, \dots, X_n$ of size $n$
from the population.
The term 'standard error' usually refers to the the standard deviation
of an estimator. In the current situation the standard error of $\bar X$
is $\sigma/\sqrt{n},$ where $\sigma$ is the standard deviation of the population.
However, if $\sigma$ is unknown and estimated by the sample standard deviation
$S= \sqrt{\frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2},$ then
$T = \frac{\bar X = \mu}{S/\sqrt{n}}$ has Student's t distribution with
$\nu = n-1$ degrees of freedom, $\mathsf{T}(\nu).$
This fact is used to make a 95% confidence for (estimating) $\mu$ of the form
$\bar X \pm t^*\frac{S}{\sqrt{n}},$ where $t^*$ cuts probability $0.025= 2.5\%$ from
the upper tail of the distribution $\mathsf{T}(\nu).$ [By the symmetry of the t distribution $-t^*$ cuts 2.5% of the probability from the lower tail of the distribution, leaving 95% in the middle.]
In this situation
the estimated standard deviation of $\bar X$ is $S/\sqrt{n}.$ Usually
the word estimated is left out because as soon as you see $S$ you know
it's an estimate of $\sigma.$
You might say you have two standard errors
of the sample mean $\bar X,$ the theoretical one $\sigma/\sqrt{n},$ and the
estimated one, $S/\sqrt{n}.$ However, if $\sigma$ is unknown, only the (estimated)
standard error $S/\sqrt{n}$ is of practical use---in making a confidence
interval as above or in testing a hypothesis about $\mu.$
Note: There are situations in which $\mu$ is unknown and $\sigma$ is known.
Then a 95% confidence interval for $\mu$ is $\bar X \pm 1.96 \frac{\sigma}{\sqrt{n}},$
where 1.96 cuts probability 2.5% from the upper tail of the standard normal
distribution. (This confidence interval is based on the fact that
$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}}$ has a standard normal distribution.)
Notice that the theoretical standard error $\sigma/\sqrt{n}$ is used when $\sigma$ is known.
Best Answer
Using sample standard deviation provides an estimate for the standard error of a sampling distribution. As you note, different samples will most likely have different sample standard deviations, leading to different standard error estimates. None of these are the standard error since none of them come from the true population standard deviation.
The bigger question is, what are you intending to do with these multiple samples? That will direct if there is more for you to do and what.