Why does variance square the deviations instead of just averaging how far each value is from the mean?

Variance squares the deviations because the signed deviations from the mean always cancel to zero, and squaring removes the sign in a form that can be expanded and combined. It is not merely a trick to hide the sign: the square is what makes the variance computable in one pass and what gives it the scaling and addition rules used everywhere else.

The definition is an average of squared distances from the expectation, not an average of distances. The square root brings the result back to the original units, which is why the standard deviation is reported and the variance is calculated with.

Take three equally likely values 0, 2 and 10. The expectation is $(0+2+10)/3 = 4$. The deviations are $-4$, $-2$ and $+6$, and their average is $(-4-2+6)/3 = 0$. That is not an accident of this example. Since $\mu$ is the weighted balance point, positive and negative deviations cancel exactly for every distribution.

A spread measure therefore has to discard the sign before averaging. Taking absolute values gives 4, 2 and 6, whose average is $(4+2+6)/3 = 4$. Taking squares gives 16, 4 and 36, whose average is $(16+4+36)/3 = 56/3$, about 18.67. The square root of $56/3$ is about 4.32, which is in the original units and is the standard deviation.

The expansion is not decoration. It turns a two-pass calculation into a one-pass calculation: keep a running total of the values and a running total of their squares, then subtract the square of the average from the average of the squares. For the three values above, the average of the squares is $(0+4+100)/3 = 104/3$, the square of the average is $4^2 = 16 = 48/3$, and the difference is $104/3 - 48/3 = 56/3$, the same variance. That is how the variance is actually computed.

Squares also give clean rules. Shifting every value by a constant leaves deviations unchanged, so variance is unchanged. Multiplying every value by a constant $a$ multiplies every deviation by $a$, so the variance is multiplied by $a^2$. For 0, 2 and 10 the variance is $56/3$; doubling to 0, 4 and 20 gives deviations $-8$, $-4$ and $12$, squared deviations 64, 16 and 144, and variance $(64+16+144)/3 = 224/3$, which is 4 times $56/3$. For unrelated quantities, variances add: $\operatorname{Var}[X+Y]=\operatorname{Var}[X]+\operatorname{Var}[Y]$. If they are related, a covariance term appears.

Mean absolute deviation is not wrong. It answers a different question: what is the typical distance from the centre, with every unit of distance counted equally. Variance answers a question in which distance is squared, so a value ten units out contributes one hundred times as much as a value one unit out. If the far outcome is merely rare but not especially costly, that weighting can be misleading. If it is a service outage, a financial loss or an engineering failure, the weighting is often exactly right.

Variance also needs a finite second moment. A distribution with sufficiently heavy tails can have an infinite variance while its mean absolute deviation is finite, and then the standard deviation is not a useful summary at all. The familiar statements that about two thirds of the weight lies within one standard deviation are statements about the bell shape, not about variance in general. The universal bound is much weaker: the weight more than $k$ standard deviations from the mean is at most $1/k^2$, so more than two standard deviations holds at most $1/4$ of the weight and more than five at most $1/25$. For a heavy tailed quantity those bounds may be the only honest error bar available.