Q:

Is standard deviation totally wrong? How can you calculate std for heights, counts and etc (positive numbers)?

Let's say I am calculating heights (in cm) and the numbers must be higher than zero.
Here is the sample list:
0.77132064
0.02075195
0.63364823
0.74880388
0.49850701
0.22479665
0.19806286
0.76053071
0.16911084
0.08833981

Mean: 0.41138725956196015
Std: 0.2860541519582141

In this example, according to the normal distribution, 99.7% of the values must be between ±3 times the standard deviation from the mean. However, even twice the standard deviation becomes negative:
-2 x std calculation = 0.41138725956196015 - 0.2860541519582141 x 2 = -0,160721044354468

However, my numbers must be positive. So they must be above 0. I can ignore negative numbers but I doubt this is the correct way to calculate probabilities using standard deviation.
Can someone help me to understand if I am using this in correct way? Or do I need to chose a different method?
Well to be honest, math is math. It doesn't matter if it is normal distribution or not. If it works with unsigned numbers, it should work with positive numbers as well! Am I wrong?
EDIT1: Added histogram
To be more clear, I have added my real data's histogram

EDIT2: Some values
Mean: 0.007041500928135767
Percentile 50: 0.0052000000000000934
Percentile 90: 0.015500000000000047
Std: 0.0063790857035425025
Var: 4.06873389299246e-05

A:

If your numbers can only be positive, then modeling them as a normal distribution may not be desirable depending on your use case, because the normal distribution is supported on all real numbers.
Perhaps you would want to model height as an exponential distribution, or maybe a truncated normal distribution?
EDIT: After seeing your data, it really looks like it might fit an exponential distribution well! You could estimate the $ \lambda $ parameter by taking, for example, a maximum likelihood approach.

A:

"What is the correct way to apply 68-95-99.7 to my case?"
One should only expect that rule of thumb for the coverage to apply exactly only if you are (1) looking at the entire (infinite) population or theoretical probability distribution, and (2) the distribution is exactly normal.
If you take a random sample of size 20, even from a genuinely normal distribution, you won't always find that 95% of the data (19 of the 20 items) lies within 2 (or 1.960) standard deviations of the mean. In fact it is neither guaranteed that 19 of the 20 items will lie within 1.960 population standard deviations of the population mean, nor that 19 of the 20 items lie within 1.960 sample standard deviations of the sample mean.
If you take a sample of data from a distribution that is not quite normally distributed, then again one would not expect the 68-95-99.7 rule to apply exactly. But it may come reasonably close to doing so, particularly if the sample size is large (the "99.7% coverage" rule-of-thumb may not be especially meaningful with a sample size below 1000) and the distribution is reasonably close to normality. In theory lots of data such as height or weight could not come from a precisely normal distribution or that would imply a small, but non-zero, probability of them being negative. Nevertheless, for data with an approximately symmetrical and unimodal distribution, where middling values are more common and extremely high or low values drop off in probability, the model of a normal distribution may be adequate for practical purposes. Incidentally you may be interested in If my histogram shows a bell-shaped curve, can I say my data is normally distributed? 
If you want theoretically binding bounds that apply to any distribution, then see Chebyshev's inequality, which states that at most $1/k^2$ of the values can lie more than $k$ standard deviations from the mean. This guarantees that at least 75% of data lie within two standard deviations of the mean, and 89% within three standard deviations. But those figures are just the theoretically-guaranteed minimum. For many roughly bell-shaped distributions, you will find that the two-standard deviation coverage figure comes much closer to 95% than to 75%, and so the "rule of thumb" from the normal distribution is still useful. On the other hand, if your data come from a distribution that is nowhere near bell-shaped, you may be able to find an alternative model that describes the data better and has a different coverage rule.
(One thing that is nice about the 68-95-99.7 rule is that it applies to any normal distribution, regardless of its parameters for mean or standard deviation. Similarly, Chebyshev's inequality applies regardless of the parameters, or even the distribution, though only gives lower bounds for coverage. But if you apply, for example, a truncated normal or skew normal model, then there isn't a simple equivalent of "68-95-99.7" coverage, because it would depend upon the parameters of the distribution.)

A:

Can someone help me to understand if i am using this in correct way?

Oh, that's easy. No, you're not using it correctly.
First off, you're using a rather small data set. Trying to tease out statistical behavior from this size set is certainly possible, but the confidence bounds are (ahem) rather large. For small data sets, deviations from the expected distributions are par for the course, and the smaller the set the greater the problem. Remember, "The Law of Averages not only permits the most outrageous coincidences, it requires them."
Worse, the particular data set you're using simply doesn't look much like a normal distribution. Think about it - with a mean of .498 you've got two samples below 0.1, and three more at .748 or above. Then you've got a cluster of 3 points between .17 and .22. Looking at this particular data set and arguing that it must be normal distribution is a pretty good case of Procrustean argument. Does that look like a bell curve to you? It's perfectly possible that the larger population does follow a normal, or modified normal, distribution, and a larger sample size would address the issue, but I wouldn't bet on it, particularly without knowing more about the population.
I say modified normal, since as Kevin Li has pointed out, technically a normal distribution includes all real numbers. As was also pointed out in comments to his answer, this does not prevent applying such a distribution over a limited range and getting useful results. As the saying goes, "All models are wrong. Some are useful." 
But this particular data set simply doesn't look like inferring a normal distribution (even over a limited range) is a particularly good idea. If your 10 data points looked like .275,.325,.375,.425,.475,.525,.575,.625,.675,.725 (mean of 0.500), would you assume a normal distribution?

