As your research shows, most discussion of repeating decimals focuses
on the simple repeating block of digits in the decimal expansion.
There is some interest in "cycling" numbers, but those are really just
the same repeating block starting at a different point:

\begin{align}
1/7 &= 0.\overline{142857} \\
&= 0.142857\overline{142857} \\
&= 0.14\overline{285714}.
\end{align}

The simple repeating block of $1/7$ is easily seen if you try to compute
$1/7$ by long division:

$$\require{enclose}
\begin{array}{rll}
0.142857 \\[-3pt]
7 \enclose{longdiv}{1.000000} \\[-3pt]
\underline{7}\phantom{00000} \\[-3pt]
30\phantom{0000} \\[-3pt]
\underline{28}\phantom{0000} \\[-3pt]
20\phantom{000} \\[-3pt]
\underline{14}\phantom{000} \\[-3pt]
60\phantom{00} \\[-3pt]
\underline{56}\phantom{00} \\[-3pt]
40\phantom{0} \\[-3pt]
\underline{35}\phantom{0} \\[-3pt]
50 \\[-3pt]
\underline{49} \\[-3pt]
1
\end{array}
$$

Once you find a remainder of $1$ after one of the subtraction steps,
you know the pattern will repeat.
(In general, for $1/n$, you don't necessarily ever find a remainder of $1$;
but as soon as you see *any* remainder that you have seen before,
you have found a repeating pattern.
For example, $1/6$ has the repeating remainder $4$.)

But long division also reveals shorter patterns that "repeat" in the
form of geometric series:

\begin{array}{rll}
0.14 \\[-3pt]
7 \enclose{longdiv}{1.00} \\[-3pt]
\underline{7}\phantom{0} \\[-3pt]
30\\[-3pt]
\underline{28} \\[-3pt]
2 \\[-3pt]
\end{array}

The remainder of $2$ indicates that the rest of the digits of $1/7$
will just be the digits of $2/7$ shifted two places to the right;
and the digits that occur there will just be the result of doubling
all the digits of $1/7$. In more algebraic notation, what happens is

\begin{align}
1 &= 7 \times 0.14 + 0.02 \\
&= 7 \times 0.14 + 0.02(7 \times 0.14 + 0.02) \\
&= 7 \times 0.14 + 0.02(7 \times 0.14 + 0.02(7 \times 0.14 + 0.02)) \\
&= 7 \times 0.14 + 0.02(7 \times 0.14 +
0.02(7 \times 0.14 + 0.02(7 \times 0.14 + 0.02))).
\end{align}

If we rewrite these same equations with all the multiplications fully
distributed over the terms in parentheses, the result is

\begin{align}
1 &= 7 \times 0.14 + 0.02 \\
&= 7 \times 0.14 + 7 \times 0.02 \times 0.14 + 0.02^2 \\
&= 7 \times 0.14 + 7 \times 0.02 \times 0.14
+ 7 \times 0.02^2 \times 0.14 + 0.02^3 \\
&= 7 \times 0.14 + 7 \times 0.02 \times 0.14
+ 7 \times 0.02^2 \times 0.14
+ 7 \times 0.02^3 \times 0.14 + 0.02^4.
\end{align}

As we continue writing equations in this pattern, we generate the terms
of the infinite geometric series

$$ \sum_{k=0}^\infty 7 \times 0.14 \times 0.02^k. $$

The method by which we develop this series suggests intuitively
that the limit of the sum as the number of terms goes to infinity is $1$,
but we can also use the known formula for the sum of a geometric series,
$ \sum_{k=0}^\infty r^k = \frac{1}{1 - r}$, to prove it:

$$ \sum_{k=0}^\infty 7 \times 0.14 \times 0.02^k
= 7 \times 0.14 \times \sum_{k=0}^\infty 0.02^k
= 0.98 \times \left( \frac{1}{1 - 0.02} \right) = 1. $$

So we have

\begin{align}
\frac17 = \frac17 \times 1
&= \frac17 \sum_{k=0}^\infty 7 \times 0.14 \times 0.02^k \\
&= \sum_{k=0}^\infty \frac{14}{100} \times \left( \frac{2}{100} \right)^k \\
&= \sum_{k=0}^\infty \frac{14 \times 2^k}{100^{k+1}} .
\end{align}

The factor of $2^k$ in each term of the sum produces the doubling phenomenon:

\begin{align}
1/7 = & \phantom{+0} 0.14 \\
& + 0.0028 \\
& + 0.000056 \\
& + 0.00000112 \\
& + 0.0000000224 \\
& + \cdots .
\end{align}

Of course, unlike the simple repeating block $142857$, after a few iterations
these doubling blocks of digits start to "overlap", meaning you actually have
to add them together rather than simply concatenating them.
You already knew that, of course, but I think it's important to emphasize that
these sequences and the simple repeating sequences come to the same
answer in *different ways*.

Notice also that $0.02^3 = 8 \times 10^{-6}$.
This would correspond a remainder of $8$ at the end of a step of long-division,
which we do not allow when dividing by $7$.
In effect, the long division algorithm uses the fact that
$$7 \times 0.02^2 \times 0.14 + 0.02^3
= 7 \times 0.000056 + 0.000008
= 7 \times 0.000057 + 0.000001$$
to stop the "doubling" effect and start from $1$ again.

You can also get a tripling effect in $1/7$ from the first step of the
long division,

\begin{array}{rll}
0.1 \\[-3pt]
7 \enclose{longdiv}{1.0} \\[-3pt]
\underline{7} \\[-3pt]
3
\end{array}

This corresponds to the equation

$$ 1 = 7 \times 0.1 + 0.3, $$

which can be further expanded to $7 \times 0.1 + 0.3(7 \times 0.1 + 0.3)$,
$7 \times 0.1 + 0.3(7 \times 0.1 + 0.3(7 \times 0.1 + 0.3))$,
and so forth, which generates the series

$$\sum_{k=0}^\infty \frac{3^k}{10^{k+1}}
= 0.1 + 0.03 + 0.009 + 0.0027 + 0.0081 + \cdots.$$

So long division is one way to discover a doubling or tripling pattern applied
to some digits (often many fewer digits than the simple repeated block contains), and there can be more than one such pattern that can
be discovered this way.

The last example (the tripling pattern of $1/7$)
is an example of the infinite sum
$\sum_{k=1}^{\infty} \frac{\left(10^x - n\right)^{k-1}}{\left(10^x\right)^k}$
which you mentioned in a comment, with $n=7$ and $x=1$.
This works as a representation of the fraction $1/n$ because
the sum is a geometric series, and

$$\sum_{k=1}^{\infty} \frac{\left(10^x - n\right)^{k-1}}{\left(10^x\right)^k}
= \frac 1n.$$

(This was shown in https://math.stackexchange.com/a/1372077/139123 using
a generalization of this formula, with $N$ instead of $10^x$
and $p$ instead of $n$.)

Let's generalize this formula in a slightly different way:

$$\sum_{k=0}^{\infty}
\frac{m\left(10^x - mn\right)^k}{\left(10^x\right)^{k+1}}.$$

Notice that this starts $k$ at $0$ instead of $1$ without changing
any of the terms of the series.
We can evaluate this as follows:

\begin{align}
\sum_{k=0}^{\infty}
\frac{m\left(10^x - mn\right)^k}{\left(10^x\right)^{k+1}}
& = \frac m{10^x} \sum_{k=0}^{\infty} \left(1 - \frac{mn}{10^x}\right)^k
\tag{A} \\
& = \frac m{10^x} \left(\frac{1}{1 - \left(1 - \frac{mn}{10^x}\right)}\right)
\tag{B} \\
& = \frac m{10^x} \left(\frac{10^x}{mn}\right)\\
& = \frac 1n.
\end{align}

Once again, we use the formula for the sum of an infinite
geometric series to get from step (A) to step (B).

This gives you a lot of flexibility in setting up geometric series to
find doubling, tripling, or other $N$-tupling digit sequences in a
repeating decimal fraction.
For example, setting $n = 7$, $m=14$, $x=2$,

$$
\frac 17
= \sum_{k=0}^{\infty} \frac{m\left(10^x - mn\right)^k}{\left(10^x\right)^{k+1}}
= \sum_{k=0}^{\infty}
\frac{14\left(10^2 - 14\times7\right)^k}{\left(10^2\right)^{k+1}}
= \sum_{k=0}^{\infty}
\frac{14 \times 2^k}{100^{k+1}}
$$
just as we found before.
Other examples are:

\begin{array}{rrrcl}
n & m & x && \qquad\text{decimal expansion} \\ \hline
19 & 5 & 2 &&
\displaystyle{\frac{1}{19}
= \sum_{k=0}^{\infty} \frac{5 \times 5^k}{100^{k+1}}}
= 0.05 + 0.0025 + 0.000125 + \cdots \\
47 & 2 & 2 &&
\displaystyle{\frac{1}{47}
= \sum_{k=0}^{\infty} \frac{2 \times 6^k}{100^{k+1}}}
= 0.02 + 0.0012 + 0.000072 + \cdots \\
49 & 2 & 2 &&
\displaystyle{\frac{1}{47}
= \sum_{k=0}^{\infty} \frac{2 \times 2^k}{100^{k+1}}}
= 0.02 + 0.0004 + 0.000008 + \cdots \\
71 & 14 & 3 &&
\displaystyle{\frac{1}{49}
= \sum_{k=0}^{\infty} \frac{14 \times 6^k}{1000^{k+1}}}
= 0.014 + 0.000084 + 0.000000504 + \cdots \\
83 & 12 & 3 &&
\displaystyle{\frac{1}{49}
= \sum_{k=0}^{\infty} \frac{12 \times 4^k}{1000^{k+1}}}
= 0.012 + 0.000048 + 0.000000192 + \cdots \\
97 & 1 & 2 &&
\displaystyle{\frac{1}{97}
= \sum_{k=0}^{\infty} \frac{1 \times 3^k}{100^{k+1}}}
= 0.01 + 0.0003 + 0.000009 + \cdots \\
199 & 5 & 3 &&
\displaystyle{\frac{1}{199}
= \sum_{k=0}^{\infty} \frac{5 \times 5^k}{1000^{k+1}}}
= 0.005 + 0.000025 + 0.000000125 + \cdots
\end{array}

There is a doubling pattern when the summation has $2^k$ in the numerator,
tripling when the summation has $3^k$ in the numerator, etc.

The procedure for generating these examples was simply: given $n$,
choose $x$ and set $m = \left\lfloor \frac{10^x}{n} \right\rfloor$.
Whether the result is an "interesting" pattern depends partly on how
hard you want to look at the digits of the decimal fraction;
for example, with $1/7$ there is
a pattern with $x=1$, $m=1$ (powers of $10-7 = 3$),
a pattern with $x=3$, $m=142$ (powers of $1000-7\times142= 6)$,
and even patterns with $x=4$ and $x=5$, but the easiest pattern to see
(in my opinion) is for $x=2$ (powers of $2$).
The effect is enhanced when $10^x - mn$ is small.
It also seems desirable for $x$ to be a fraction of the period of the repeating decimal, and not very large.
You can find $10^x - mn$ for as many values of $x$ as you wish
by starting with $x=0$ (in which case $10^x - mn = 1$),
and for each successive value of $x$ ($x = 1,2,3,\ldots$),
multiply the previous result by $10$ and find the remainder
when you divide by $n$.

It does not seem simple to predict which $n$ will have small enough values
of $10^x - mn$ for small enough values of $x$
without going through this exercise (even setting aside the fact that
"small enough" is a subjective measurement here).
But by setting this up in a spreadsheet I found some additional
"interesting" patterns for $n < 200$:

$$
\frac{1}{17} = \sum_{k=0}^{\infty} \frac{588 \times 4^k}{10000^{k+1}},
\quad
\frac{1}{43} = \sum_{k=0}^{\infty} \frac{232558 \times 6^k}{10000000^{k+1}},
\quad
\frac{1}{51} = \sum_{k=0}^{\infty} \frac{196 \times 4^k}{10000^{k+1}},
$$
$$
\frac{1}{119} = \sum_{k=0}^{\infty} \frac{84 \times 4^k}{10000^{k+1}},
\quad
\frac{1}{127} = \sum_{k=0}^{\infty} \frac{7874 \times 2^k}{1000000^{k+1}},
\quad
\frac{1}{167} = \sum_{k=0}^{\infty} \frac{5988 \times 4^k}{1000000^{k+1}}.
$$

And of these, I would not consider $1/17$ or $1/43$ to be "easy" to see
even after you know what to look for.

## Best Answer

If I understood correctly, your question is about why certain groups of digits in the decimal expansion of $1/7$ are multiples of $7$, and why they also have a factor of $2^n$.

For a short version: jump to the end of my answer, the following is an explanation which is not as clean as the last paragraph.You achieved to express $1/7$ as a geometric series

$$7\cdot\sum_{n=1}^{\infty} \left(\frac2{100}\right)^n.$$

The $100$ is not very surprising, as we are talking about digits in the

decimal system. So powers of $10$ are always present. What might be puzzling is where the $2$ comes from. The limit of the series is$$7\cdot\frac{2/100}{1-2/100}=7\cdot\frac2{100-2}=7\cdot\frac2{98}$$

and the reason this perfectly works out to $1/7$ is because $98=\color{red}{2}\cdot7\cdot 7$. The reason why this works so perfectly for $7$ (and not for other numbers $-$ then it would not really be surprising), is that $7^2+1=50$ is exactly $10^{2}/{\color{red}{2}}$ (and this is where $2$ comes into the game).

Let me explain. If we want that groups of digits of $1/p$ such that they are multiples of $p$ itself, we have to express it as a geometric series

$$\sum_{n=1}^{\infty}\left[p\cdot\left(\frac{q}{10^k}\right)^n\right].$$

The $10^k$ is an artifact of the decimal system and makes that our digits are (to some extent) exactly of the form $p\cdot q^n$ (e.g. $7\cdot2^n$ in your case). There are some ways to tweak this system, but let's stick to it for now. Evaluating this series gives

$$p\cdot\frac{q/10^k}{1-q/10^k}=p\cdot\frac{q}{10^k-q}.$$

For this to perfectly be $1/p$, we need $10^k-q=q\cdot p^2$ (check it, everything cancels out). We rearrange this to

$$q=\frac{10^k}{p^2+1}.$$

Here you see why this works for $p=7$. Then it turns out we can choose $k=2$ to get $q=2$. Until now I found no other number for which we have $p^2+1$ in such a nice form. If you allow that the digit groups are of the form $\alpha p\cdot q^n$ with some additional factor $\alpha$, then we are looking for combinations in which $\alpha p^2+1$ divides $10^k$. It is still not easy to find such combinations, but I found one or two nicer ones.

Using my method, I found that

$$\frac1{127}\approx 0.\color{lightgray}{00}\color{red}{7874}\color{lightgray}{0}15748\color{lightgray}{0}\color{red}{31496}\color{lightgray}{0}62992\color{red}{125984}251\, ...$$

which turned out to be generated by

\begin{align} \color{lightgray}{00}\color{red}{7874} &= 127 \cdot 31 \cdot 2^1 \\ \color{lightgray}{0}15748 &= 127 \cdot 31 \cdot 2^2 \\ \color{lightgray}{0}\color{red}{31496} &= 127 \cdot 31 \cdot 2^3 \\ \color{lightgray}{0}62992 &= 127 \cdot 31 \cdot 2^4 \\ \color{red}{125984} &= 127 \cdot 31 \cdot 2^5 \\ \cdots \end{align}

This example uses $q=2$, $k=6$, $p=127$ and $\alpha=31$. The reason for this nice pattern (and the occurence of $\color{red}2^n$) is again that $127^2\cdot31+1=500000=10^6/\color{red}2$.

When you are less restrictive on this kind of occuring pattern, e.g. $p$ needs no longer to be part of the digit sequence of $1/p$, then we can build an even nicer example from the one above:

$$\frac1{127^2}\approx 0.\color{lightgray}{0000}62\color{lightgray}{000}124\color{lightgray}{000}248\color{lightgray}{000}496\color{lightgray}{000}992\color{lightgray}{00}1984\color{lightgray}{00}3968\,...$$

where the black digits are just $31\cdot 2^n$.

Another

crazyexample is$$\frac1{17}\approx 0.\color{lightgray}0\color{red}{58823529411764705882352941176470588235294}11764705\, ...$$

which follows the pattern $a(n)=17\cdot 1730103806228373702422145328719723183391\cdot 2^n$. For small values of $n$ we find

\begin{align} a(1)&=\color{lightgray}0\color{red}{58823529411764705882352941176470588235294}\\ a(2)&=117647058823529411764705882352941176470588\\ \cdots \end{align}

as expected. Unfortunately the numbers are so long that I cannot really show you how perfectly it fits. This pattern uses $p=17$, $q=2$, $k=42$ and an enormous $\alpha$ seen above. Once again we observe $17^2\cdot \alpha+1=10^{42}/\color{red}2$.

A more technical observationThinking about the problem so much brought me to an even more direct insight.

Proof.$$\frac1a=b\cdot\frac1{ab}=b\cdot\frac1{10^k/q-1}=b\cdot\frac{q/10^k}{1-q/10^k}=\sum_{n=1}^{\infty} \left[b\cdot \left(\frac q{10^k}\right)^n\right].\qquad\square$$

This was used above and can be seen in all of the examples, e.g.

$$7\cdot7=\frac{10^2}2-1,\qquad 127^2\cdot 31=\frac{10^6}2-1$$

This shows a nice duality. Because of the symmetry between $a$ and $b$ we also see that $1/31$ contains the pattern $127^2\cdot 2^n$. No such duality can be observed in $1/7$ because this was the one case with $a=b=7$. We also see that the only possible values for $q$ are $2^s5^r$ because these are the only possible divisors of $10^k$.

This final observation allows us to generate an infinitude of examples with nice digit patterns. So I will simply stop here despite the problem really catching my attention. The examples presented above are not so nice in this new light but you can certainly find better ones now.