I know that 95% of the observations under a normal distribution fall approximately under 2 standard deviations from the mean.
1.96, but yes.
Does this change when the distribution has fat tails?
Yes. It also changes with thin tails.
However, there are a lot of distributions with very roughly 95% within two standard deviations. Indeed, if we look at continuous symmetric unimodal distributions whose variance exists, there must be between 88% and 100% within two standard deviations.
On the other hand, in the general case, the limit is given by the Chebyshev inequality - i.e. it may be as low as 3/4.
What is the name of the theorem that guarantees this?
You don't actually need a theorem; a counterexample to the assertion that it's the case for all distributions would suffice to say that it changes (but since the Chebyshev bound is achievable, it's a good one to look at if you want to mention something). All you really need to do is just compute it for a few different distributions. e.g. look at a uniform and a 
, and some asymmetric and discrete cases.
One interesting case to consider is a distribution that has probability of 
at 
and 
at 0. Now move 
up and down between 0 and 1 and see the fraction inside two standard deviations can be changed quite a lot.
In that example, the variance is 
. If 
then the proportion inside 2 s.d.s is 
(so that proportion can be any value in 
). Then for 
the proportion jumps to exactly 1, which means we can demonstrably achieve any value in 
, and 0.75 is the Chebyshev bound, so we can't go below 0.75. (It's a very handy distribution to play with.)
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I know that 95% of the observations under a normal distribution fall approximately under 2 standard deviations from the mean.
1.96, but yes.
Does this change when the distribution has fat tails?
Yes. It also changes with thin tails.
However, there are a lot of distributions with very roughly 95% within two standard deviations. Indeed, if we look at continuous symmetric unimodal distributions whose variance exists, there must be between 88% and 100% within two standard deviations.
On the other hand, in the general case, the limit is given by the Chebyshev inequality - i.e. it may be as low as 3/4.
What is the name of the theorem that guarantees this?
You don't actually need a theorem; a counterexample to the assertion that it's the case for all distributions would suffice to say that it changes (but since the Chebyshev bound is achievable, it's a good one to look at if you want to mention something). All you really need to do is just compute it for a few different distributions. e.g. look at a uniform and a , and some asymmetric and discrete cases.
One interesting case to consider is a distribution that has probability of at and at 0. Now move up and down between 0 and 1 and see the fraction inside two standard deviations can be changed quite a lot.
In that example, the variance is . If then the proportion inside 2 s.d.s is (so that proportion can be any value in ). Then for the proportion jumps to exactly 1, which means we can demonstrably achieve any value in , and 0.75 is the Chebyshev bound, so we can't go below 0.75. (It's a very handy distribution to play with.)
I understand standard deviation a bit but not fully so kindly someone explain two standard deviation.
Compilation and expansion of comments:
Let's presume your data is Normally distributed.
If you want to form two-sided error bars (or confidence intervals), say at the 95% level, you will need to base that on the Student t distribution with n-1 degrees of freedom, where n is the number of data points. You propose to have 2 data points, therefore requiring use of Student t with 1 degree of freedom.
95% 2-sided error bars for n = 2 data points require a multiplicative factor of 12.71 on the sample standard deviation, not the familiar factor of 1.96 based on the Normal (Student t with 
degrees of freedom). The corresponding multiplicative factor for n = 3 data points is 4.30.
The situation gets even more extreme for two-sided 99% error bars (confidence intervals).
As you can see, at either confidence level, there's a big "savings" in the multiplicative factor if you have 3 data points instead of 2. And you don't get dinged as badly by the use of n-1 vs. n in the denominator of sample standard deviation.
n Confidence Level Multiplicative Factor
2 0.95 12.71
3 0.95 4.30
4 0.95 3.18
5 0.95 2.78
infinity 0.95 1.96
2 0.99 63.66
3 0.99 9.92
4 0.99 5.84
5 0.99 4.60
infinity 0.99 2.58
Setting aside your initial explanation of the time-series context, it might be useful to look at this as a simple case of observing two data points. For any two observed values 
the sample standard deviation is 
. This statistic is exactly as informative as giving the sample range of the two values (since it is just a scalar multiple of that statistic). There is nothing inherently wrong with using this statistic as information on the standard deviation of the underlying distribution, but obviously there is a great deal of variability to this statistic.
The sampling distribution of the sample standard deviation depends on the underlying distribution for the observable values. In the special case where 
are normal values you have 
which is a scaled half-normal distribution. Obviously this means that your sample standard deviation is quite a poor estimator of the standard deviation parameter (biased and with high variance), but that is to be expected with so little data.