The distribution of the difference (sample.mean-population.mean) depends on the population standard deviation and the sample size (in particular, the standard deviation of the difference is related to both -- it's $\sigma/\sqrt{n}$).
This is the (true) standard error of the sample mean.
The distribution of percentage difference will depend on where the population mean is.
Consider that percentage is basically dividing the raw difference by the mean and multiplying by 100. (I'll leave aside the x100 for now.)
So instead of $\sigma/\sqrt{n}$ for the standard deviation of the difference, you're dealing with $\frac{\sigma}{\mu\sqrt{n}}$ for the standard deviation of the relative difference.
Consider I have a mean of 100, and and a standard error of 5. I compute a relative standard error of 0.05 (5%). Now if I subtract 90 from every observation, my standard error is unchanged (it doesn't involve the mean), but my relative standard error has jumped to 50. So I can't make general comments about the size of the relative standard error without reference to where the mean is. It depends on that mean, quite directly.
This is a warning -- when you do simulation, you can't just make conclusions based on one set of parameter values unless you really understand how it will generalize to other values. If you don't, you have to let the simulation tell you how things change as you play with mean and standard deviation independently.
Now $\frac{\sigma}{\mu}$ is called the coefficient of variation. It's often useful in situations where spread tends to increase proportionally when the mean increases (generally the same time that percentage changes make the most sense, basically).
The CLT (at least in some of its various forms) tells us that in the limit as $n\to\infty$ distribution of a single standardized sample mean ($\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}$) converges to a normal distribution (under some conditions).
The CLT does not tell us what happens at $n=50$ or $n=50,000$.
But in attempting to motivate the CLT, particularly when no proof of the CLT is offered, some people rely on the sampling distribution of $\bar{X}$ for finite samples and show that as larger samples are taken that the sampling distribution gets closer to the normal.
Strictly speaking this isn't demonstrating the CLT, it's nearer to demonstrating the Berry-Esseen theorem, since it demonstrates something about the rate at which the approach to normality comes in -- but that in turn would lead us to the CLT, so it serves well enough as motivation (and in fact, often something like the Berry-Esseen comes closer to what people actually want to use in finite samples anyway, so that motivation may in some sense be more useful in practice than the central limit theorem itself).
the distribution of these sample means would be normal.
Well, no, they would be non-normal but they would in practice be very close to normal (heights are somewhat skew but not very skew).
[Note again that the CLT really tells us nothing about the behavior of sample means for $n=50$; this is what I was getting at with my earlier discussion of Berry-Esseen, which does deal with how far from a normal cdf the distribution function of standardized means can be for finite samples]
The real world case I am thinking about is doing statistics on a dataset of 50,000 twitter users. That dataset obviously isn't repeated samples, it is just one big sample of 50,000.
For many distributions, a sample mean of 50,000 items would have very close to a normal distribution -- but it's not guaranteed, even at n=50,000 that you will have very close to a normal distribution (if the distribution of the individual items is sufficiently skewed, for example, then the distribution of sample means may still be skew enough to make a normal approximation untenable).
(The Berry-Esseen theorem would lead us to anticipate that exactly that problem might occur -- and demonstrably, it does. It's easy to give examples to which the CLT applies but for which n=50,000 is not nearly a large enough sample for the standardized sample mean to be close to normal.)
Best Answer
I think you might be confusing the expected sampling distribution of a mean (which we would calculate based on a single sample) with the (usually hypothetical) process of simulating what would happen if we did repeatedly sample from the same population multiple times.
For any given sample size (even n = 2) we would say that the sample mean (from the two people) estimates the population mean. But the estimation accuracy -- that is, how good a job we've done of estimating the population mean based on our sample data, as reflected in the standard error of the mean -- will be poorer than if we had a 20 or 200 people in our sample. This is relatively intuitive (larger samples give better estimation accuracy).
We would then use the standard error to calculate a confidence interval, which (in this case) is based around the Normal distribution (we'd probably use the t-distribution in small samples since the standard deviation of the population is often underestimated in a small sample, leading to overly optimistic standard errors.)
In answer to your last question, no we don't always need a Normally distributed population to apply these estimation methods -- the central limit theorem indicates that the sampling distribution of a mean (estimated, again, from a single sample) will tend to follow a normal distribution even when the underlying population has a non-Normal distribution. This is usually appropriate for "bigger" sample sizes.
Having said that, when you have a non-Normal population that you're sampling from, the mean might not be an appropriate summary statistic, even if the sampling distribution for that mean could be considered reliable.