Let $Y=X^2$. Then $Y$ takes on the values $1$, $4$, and $9$ respectively when $X$ takes on the values $1$, $2$, and $3$.
Thus $p_Y(1)=p_X(1)$, $p_Y(4)=p_X(2)$, and $p_Y(9)=p_X(3)$.
Now for calculating $E(Y)$ we just use the formula the post started with, namely
In our case, we get $1\cdot p_Y(1)+4\cdot p_Y(4)+9\cdot p_Y(9)$. Equivalently, $E(Y)= 1\cdot p_X(1)+4\cdot p_X(2)+9\cdot p_X(3)$.
To answer your question more explicitly, we do not use $1^2(p_X(1))^2+2^2(p_X(2))^2+3^2(p_X(3))^2$ because, for example, $\Pr(X^2=3^2)$ is not $(\Pr(X=3))^2$. In fact, $\Pr(X^2=3^2)=\Pr(X=3)$.
As to your question about average, yes, the mean is a very important measure of average value. The only serious competitor is the median.
Mean and median can be quite different. For example, imagine a population in which a small minority is insanely rich, while the vast majority of the population is struggling. Then the mean income of the population may be substantially higher than the median income. Is either one a "better" measure of average wealth? I would argue that in this case the median is ordinarily of greater relevance. But for certain planning purposes, such as level of tax revenues, the mean may be more useful.
The mathematics of the mean is substantially simpler than the mathematics of the median. For example, the mean of a sum of two random variables is the sum of the means. The median of a sum is a far more complicated object.