First of all
$$
\frac{24}{7} = \frac{3 \cdot 7 + 3}{7} = 3 + \frac{3}{7},
$$
so let me answer why
$$
\frac{3}{7} = 0.428571\,428571\,428571\,428571\,428571 \cdots
$$
Let
$$
\begin{align}
x &= 0.428571 \\
&+ 0.000000\,428571 \\
&+ 0.000000\,000000\,428571 \\
&+ 0.000000\,000000\,000000\,428571 \\
&+ 0.000000\,000000\,000000\,000000\,428571 \\
&\phantom{\;\;}\vdots \\
&= \frac{428571}{1000000^1} + \frac{428571}{1000000^2}
+ \frac{428571}{1000000^3} + \frac{428571}{1000000^4}
+ \frac{428571}{1000000^5} +\cdots
\end{align}
$$
with each additional term being a million times smaller than the previous. Convince yourself that
$$
1\,000\,000 x = 428\,571 + x
$$
or that
$$
999\,999 x = 428\,571.
$$
So, it turns out that
$$
x = \frac{428\,571}{999\,999} = \frac{3}{7}.
$$
Why did this work? If you consider the list of numbers of the form
$$
\begin{align}
9 &= 10^1 - 1 \\
99 &= 10^2 - 1 \\
999 &= 10^3 - 1 \\
9\,999 &= 10^4 - 1 \\
99\,999 &= 10^5 - 1 \\
999\,999 &= 10^6 - 1 \\
&\vdots
\end{align}
$$
eventually $7$ will divide one of them (without remainder). Does $7$ divide $9$? No. Does $7$ divide $99$? No. $\dots$ Does $7$ divide $999\,999$ Yes!
$$
999\,999 = 7 \cdot 142\,857
$$
so
$$
\frac{3}{7} = \frac{3 \cdot 142\,857}{7 \cdot 142\,857} = \frac{428\,571}{999\,999}
$$
and the repeating block of the decimal expansion is $428571$.
For any fraction $\frac{a}{b}$, find a number of the form $10^N - 1$ ($N$ many $9$s) that is a multiple of $b$, and this is always possible! Say that $b \cdot r = 10^N - 1$ on the nose; then
$$
\frac{a}{b} = \frac{a \cdot r}{10^N - 1},
$$
so the string of $N$ digits that make up $a \cdot r$ forms the repeating block in the decimal expansion.