The standard error of the mean, what you call $\sigma_{\bar x}$, is a function of a sample drawn from the population.
Suppose we model the population as a discrete random variable $X$ with probability mass function $$\Pr[X = 2] = \Pr[X = 5] = \Pr[X = 7] = \Pr[X = 10] = \frac{1}{4}.$$ Then $\mu = \operatorname{E}[X] = 6$ and $\sigma^2 = \operatorname{Var}[X] = \frac{17}{2}$ as we would expect. Then a sample is a set of IID random variables $$(X_1, X_2, \ldots, X_n)$$ drawn from this distribution, and the sample mean and the variance of the sample mean are $$\bar X = \frac{1}{n} \sum_{i=1}^n X_i, \quad \operatorname{Var}[\bar X] \overset{\text{iid}}{=} \frac{\operatorname{Var}[X]}{n} = \frac{\sigma^2}{n}.$$ Thus the standard error of the mean is $$SEM = \sqrt{\operatorname{Var}[\bar X]} = \frac{\sigma}{\sqrt{n}}.$$ None of these formulas relies on any particular distributional assumption, only that the population mean and variance are finite.
How this applies to the above distribution is, for instance, we draw a sample of size $n = 2$; e.g., $(X_1, X_2)$ is our sample and this yields $$\operatorname{Var}\left[\frac{X_1 + X_2}{2}\right] = \frac{17}{4} \implies SEM = \frac{\sqrt{17}}{2}.$$ Such a sample is taken with replacement; the joint distribution is a $4 \times 4$ table of ordered pairs taken from $P$; e.g., $$(2,2), (2,5), (2,7), (2,10), \\ (5,2), (5,5), (5,7), (5,10) \\ (7,2), (7,5), (7,7), (7,10) \\ (10,2), (10,5), (10,7), (10,10),$$ and each such outcome has probability $1/16$. In general, a sample of size $n$ would have a joint distribution on the set of ordered $n$-tuples whose elements are drawn from $P$.
If your sample is without replacement from a finite population size $N$, then the variance of the sample mean is not $\dfrac{\sigma^2}{n}$ but $\dfrac{N-n}{N-1}\dfrac{\sigma^2}{n}$
So in your example of $n=100$, $N_1=120$ and $N_2=120000$, then instead of a variance of $\dfrac{\sigma^2}{100}$,
we would get $\dfrac{\sigma^2}{595}$ for the smaller population and about $\dfrac{\sigma^2}{100.083}$ for the variance population, with the variance for the smaller population substantially smaller, in line with your intuition
Best Answer
First of all, you are right with you answer. I think your confusion Côme from the fact you have two distinct situation.
First situation: you take sample of 25 induviduals from a population. Here $n=25$ and you used it to find $\bar X_k$.
Second situation: you have 64 of these d'amples. Here $n=64$. You used it to evalutate $\mu_{\bar X}$.
For your standard deviation $\sigma_{\bar X}$, you are dealing with values from the second situation, so $n=64$.
It is always a lot of fun to deal with those multileveled questions. It is important to label your variable to distinguish them.
EDIT OP added
Once again, it is a matter te define variable properly.
Level $0$: the population has a mean $\mu$ and a standard deviation of $\sigma$.
Level $1$: sample of size $25$. For each sample, you will have a mean of $\bar X_k$ and a standard deviation of $\sigma_k$ (the standard deviation of the sample).
The expected mean distribution has a mean of $E[\bar X_k]=\mu$ (same has population) and a standard deviation of $$\sqrt{\mathrm{Var}[\bar X_k]}=\frac{\sigma}{\sqrt{25}}$$
Level $2$: $64$ samples. You took $64$ means of samples evaluate the mean.
The expected mean of the value of the level $2$ mean is $$E[\bar X]=E[\bar X_k]=\mu$$ And the variance will be $$\sqrt{\mathrm{Var}[\bar X]}=\frac{\sqrt{\mathrm{Var}[\bar X_k]}}{\sqrt{64}}=\frac{\sigma}{\sqrt{25}\sqrt{64}}$$
TL;DR Now to answer about your $$\sigma_{\bar X}=\sqrt{\frac{\sum_{k=1}^{64}(\bar X_k -\mu_{\bar X})^2}{64}}$$ It is the standard deviation of your $64$ sample means, it refer to the expected distribution of the level $1$. So it should be around $\frac{\sigma}{\sqrt{25}}$.