How do we get the last inequality, $c_1\frac{|E|}{|x|^n} \le \chi^*_E(x) \le c_2\frac{|E|}{|x|^n} \text{ for large } |x|$? Why do we need $|x|$ to be large?
Let $x\in \mathbb R^n$. For any measurable $E$, $$\chi^*_E(x) = \sup\left\{\frac{|E\cap Q|}{|Q|}: Q \text{ has center } x\right\}.$$
If $E$ is bounded and $Q^x$ denotes the smallest cube with center $x$ containing $E$, then $$\frac{|E\cap Q^x|}{|Q^x|} = \frac{|E|}{|Q^x|}.$$
It follows that there are positive constants $c_1$ and $c_2$ such that $$c_1\frac{|E|}{|x|^n} \le \chi^*_E(x) \le c_2\frac{|E|}{|x|^n} \text{ for large } |x|.$$
I will use the notation $Q_{r}$ to denote the cube centered at $x$ and side-length $r$. If $r_{x}$ is the side-length of the smallest cube centered at $x$ and containing $E$, then for every $r>r_{x}$ we have that$$ \frac{|Q_{r}\cap E|}{|Q_{r}|}=\frac{|E|}{|Q_{r}|}=\frac{|E|}{r^{n}}\leq \frac{|E|}{r_{x}^{n}}=\frac{|E|}{|Q_{r_{x}}|}=\frac{|Q_{r_{x}}\cap E|}{|Q_{r_{x}}|}. $$ This shows that all cubes centered at $x$ and side-lenght $r>r_{x}$ are not good competitors for the supremum, that is,$$ \chi_{E}^{\ast}(x)=\sup\left\{ \frac{|Q_{r}\cap E|}{|Q_{r}|}:\,Q_{r}\text{ has center at }x\text{ and side-length }r\leq r_{x}\right\} . $$ Now let $B(0,R)$ be the smallest ball centered at the origin and containing $E$. Given $|x|>2R$, find $k\geq2$ such that $kR\leq|x|<(k+1)R$. If $$r<\frac{|x|}{4\sqrt{n}}\leq\frac{(k+1)R}{4\sqrt{n}}\leq\frac{2kR}{4\sqrt{n} }=\frac{kR}{2\sqrt{n}},$$ then for every $y\in Q_{r}$ we have that $|y-x|\leq\sqrt{n}r$ and so $|y|\geq|x|-|y-x|\geq kR-\sqrt{n}r>kR-\frac{kR}% {2}=\frac{kR}{2}\geq R$, which means that $Q_{r}$ does not intersect $B(0,R)$ and, in turn, it does not intersect $E$. Thus, when $kR\leq|x|<(k+1)R$, we can restrict our attention to cubes of side-length $r\geq\frac{|x|}{4\sqrt{n}}$. In turn, $\frac{1}{r^{n}}\leq\frac{(4\sqrt{n})^{n}}{|x|^{n}}$ and so$$ \frac{|Q_{r}\cap E|}{|Q_{r}|}\leq\frac{|E|}{|Q_{r}|}=\frac{|E|}{r^{n}}% \leq\frac{(4\sqrt{n})^{n}}{|x|^{n}}, $$ which shows that $\chi_{E}^{\ast}(x)\leq\frac{(4\sqrt{n})^{n}}{|x|^{n}}$.
On the other hand, if $r\geq4|x|$, then for every $y\in E$, $|y_{i}-x_{i}% |\leq|y_{i}|+|x_{i}|\leq R+|x|<\frac{1}{2}|x|+|x|\leq2|x|\leq\frac{r}{2}$, which shows that $E$ is contained in $Q_{r}$ and so$$ \frac{|Q_{r}\cap E|}{|Q_{r}|}=\frac{|E|}{|Q_{r}|}=\frac{|E|}{r^{n}}% $$ and $r_{x}\leq4|x|$. Hence, $$ \chi_{E}^{\ast}(x)=\sup\left\{ \frac{|Q_{r}\cap E|}{|Q_{r}|}:\,Q_{r}\text{ has center at }x\text{ and side-length }\frac{|x|}{4\sqrt{n}}\leq r\leq4|x|\right\} . $$ Now for any such $\frac{|x|}{4\sqrt{n}}\leq r\leq4|x|$, that is $\frac {1}{(4|x|)^{n}}\leq\frac{1}{r^{n}}\leq\frac{(4\sqrt{n})^{n}}{|x|^{n}}$, and so$$ \frac{|Q_{r}\cap E|}{|Q_{r}|}=\frac{|Q_{r}\cap E|}{r^{n}}\geq\frac{|Q_{r}\cap E|}{(4|x|)^{n}}. $$ In turn, \begin{align*} \chi_{E}^{\ast}(x) & \geq\frac{1}{(4|x|)^{n}}\sup\bigg\{ |Q_{r}\cap E|:\,Q_{r}\text{ has center at }x\\&\text{ and side-length }\frac{|x|}{4\sqrt{n} }\leq r\leq4|x|\bigg\} =\frac{1}{(4|x|)^{n}}|Q_{r_{x}}\cap E|=\frac{1}{(4|x|)^{n}}|E|. \end{align*} As for why do we care, Zach Boyd already gave a perfectly good answer. In particular the inequality from below shows that the maximal function is not integrable even if $\chi_E$ is.