How the step of squaring the deviations in Standard Deviation overcomes the drawback of ignoring the signs of mean deviation. 
$$\text{Mean deviation from mean}=\frac1N\sum_{i=1}^nf_i|x_i-\bar x|$$and $$\text{Standard Deviation ($\sigma$)}=\sqrt{\frac1N\sum_{i=1}^nf_i(x_i-\bar x)^2}$$
  The step of squaring the deviations in SD overcomes the drawback of ignoring the signs of mean deviation.

How is the problem overcome? In the SD too, we ignore the signs by squaring, aren't we?

The principal reason for the widespread use of mean square deviation instead of mean absolute deviation is that if random variables $X_1,\ldots,X_n$ are independent, then
$$
\operatorname{var}(X_1+\cdots+X_n)=\operatorname{var}(X_1)+\cdots+\operatorname{var}(X_n).
$$
That makes it possible to related the dispersion of the sum, and hence the average, of the members of a sample to the dispersion in the population from which the sample was taken, and in particular, we'd have no central limit theorem otherwise.  Without the central limit theorem, the usual method of finding a confidence interval for a population mean would not be justified.
Whoever told you that the reason is to overcome an alleged drawback of ignoring signs was probably confused.
