What makes the Var(x)+Var(y)=var(x+y) property important? What makes the Var(x)+Var(y)=var(x+y) property important?
It was taught in my statistics class
 A: In the 18th century Abraham de Moivre considered this problem (his account of which you can find in his book The Doctrine of Chances: If you toss a fair coin $1800$ times, what is the probability that the number of heads in in a specified interval, for example at least $880$ but no more than $906$?  In the course of solving that problem, he first introduced the "bell-shaped curve"
$$
\varphi(x) = \text{constant}\cdot e^{-x^2/2}
$$
where the "constant" is chosen so as to make this a probability density function.  (De Moivre found the constant numerically, and James Stirling later showed that it's $1/\sqrt{2\pi\,{}}$.  This is the "normal distribution" with expectation $0$ and variance $1$.  It follows that the normal density with expectation $\mu$ and variance $\sigma^2$ has density
$$
x\mapsto \text{same constant}\cdot\frac1\sigma\varphi\left(\frac{x-\mu}{\sigma}\right).
$$
De Moivre found that the distribution of the number of heads can be made as close as desired to the normal distribution with the right values of $\mu$ and $\sigma$ by making the number of tosses large enough, and $1800$ was plenty.
But how do you know the right values of $\mu$ and $\sigma$?
In the case of $\sigma$, you just use the fact that the variance of the sum of independent random variable is the sum of the variances, i.e. the property of variances that you learned in your statistics class.  Other simple measures of statistical dispersion lack that property, and so cannot be applied to this coin-toss problem.
It's used in statistics all the time.  How do you know that if you take a sample of size $n$ from a population that's not necessarily normally distributed, you can use the two numbers
$$
\bar x \pm 1.96\frac{S}{\sqrt{n}}
$$
as endpoints of a $95\%$ confidence interval for the population mean, where $S$ is the sample standard deviation?  In particular, where does "$S/\sqrt{n}$ come from?  The answer is that it comes from that same property of variances that you were taught.
