I propose to put some visuals/intuition to your question... using an empirical approach (bootstrapping) to make it more concrete, especially in reference to the following:
Usually experiments can't or just aren't repeated and only have 1 sample from a population
As you highlighted it, we are talking about the standard error of a statistic (the mean in our case). So, Let's assume that you have a random sample of 20 people's height from a given country:
## [1] 192.3214 144.4797 151.3796 155.2519 147.5844 147.9056 171.1867 159.3074
## [9] 163.0097 190.9857 165.8155 198.2192 192.2418 165.3628 186.9498 167.3355
## [17] 148.6400 156.6933 160.8472 174.4827
From this sample, you get a mean of 167 and a standard deviation of 17.
You have only one random sample, but you can imagine that if you could take another one, you might get similar values, sometimes duplicates or sometimes more extreme values... but something that will look like to your initial random sample.
So, from these initial sample values and without inventing new ones (only resampling with replacement), you can imagine many other samples. For example, we can imagine three as follows:
## [1] 165.8155 159.3074 148.6400 165.3628 155.2519 151.3796 192.2418 163.0097
## [9] 159.3074 192.2418 186.9498 163.0097 144.4797 198.2192 159.3074 190.9857
## [17] 165.3628 159.3074 167.3355 156.6933
## [1] 147.5844 147.9056 151.3796 163.0097 167.3355 159.3074 167.3355 156.6933
## [9] 156.6933 159.3074 147.9056 190.9857 192.2418 171.1867 198.2192 147.9056
## [17] 155.2519 167.3355 148.6400 165.8155
## [1] 192.2418 198.2192 156.6933 192.3214 148.6400 192.3214 198.2192 165.8155
## [9] 167.3355 144.4797 163.0097 148.6400 159.3074 163.0097 163.0097 174.4827
## [17] 165.3628 165.8155 174.4827 159.3074
Their respective mean will be different from the initial one... but what is interesting is that if we repeat this resampling exercise 10,000 times, for instance, and we calculate the mean for each of these generated samples, we will get something like that (leaving the R code here, just to illustrate it), a distribution of means centered around the initial sample mean:
set.seed(007)
spl <- 167+17*scale(rnorm(20))[,1] #Forcing to have same mean and sd for all samples
library(boot)
myFunc <- function(data, i){
return(mean(data[i]))
}
bootMean <- boot(spl , statistic=myFunc, R=10000)
hist(bootMean$t, xlim=c(150,185), main="Sample size n=20")
abline(v=mean(spl), col="blue")

So, the histogram above represents the distribution of means of 10,000 samples… that we constructed from our initial sample. Empirically, we can determine the standard deviation of this (sampling) distribution (which is our standard error of the mean):
sd(bootMean$t)
## [1] 3.74095
Interestingly enough, if we calculate the formula for the standard error $\frac{s}{\sqrt n}$, we get something very similar:
sd(spl)/sqrt(20)
## [1] 3.801316
The standard error of the mean tells us about the spread our data around the mean.
To finish this intuitive overview, let's see what happen if we increase our initial sample size (to understand the impact of this $\sqrt{n}$).


So, if we increase the sample size, the standard error gets unsurprisingly smaller... we reduce the error in estimating the population mean. Again, we can empirically see that the formula still holds:
sd(bootMean$t)
## [1] 0.7740625
sd(spl)/sqrt(500)
## [1] 0.7602631