We look at $2^n$, where $n$ ranges over the non-negative integers.
The key is the fact that $2^6$ has remainder $1$ on division by $9$. Using congruence notation, we have $2^6\equiv 1\pmod{9}$. Let $n$ be any non-negative integer. We can express $n$ as $6q+r$, where $0\le r\le 5$ ($q$ stands for quotient, $r$ for remainder).
It follows that
$$2^n=2^{6q+r}=(2^6)^q 2^r\equiv (1)^q 2^r\equiv 2^r\pmod{9}.$$
So the remainder when you divide $2^n$ by $9$ depends only on $r$. For $r=0$, $1$, $2$, $3$, $4$, and $5$, these remainders are, as you observed, $1$, $2$, $4$, $8$, $7$, and $5$.
To connect this with sums of (decimal) digits, observe that a decimal number like $6852$ is just $(6)(10^3)+(8)(10^2)+(5)(10^1) +(2)(10^0)$. But for any non-negative integer $k$, we have $10^k\equiv 1\pmod{9}$. So $6852\equiv 6+8+5+2 \pmod{9}$. Thus the remainder when $6852$ is divided by $9$ is the same as the remainder when $6+8+5+2$ is divided by $9$. Asimilar remark holds for any non-negative integer expressed in decimal form.
Since remainders when $2^n$ is divided by $9$ cycle with period $6$, and the "casting out nines" process gives us these remainders, the pattern you observed continues forever.
You extended the pattern to negative exponents. This is an interesting observation that I do not recall seeing before. Express $2^{-n}$ as a decimal, by noting that
$$2^{-n}=\frac{1}{2^n}=\frac{5^n}{10^n}.$$
Then (essentially) you looked at the digit sum (modulo $9$) of $5^n$. Modulo $9$, the numbers $5^n$ cycle with period $6$, for the same reason as with $2^n$.
Now calculate $5^n$ modulo $9$, for $n=0,1, 2, 3, 4, 5$. We get that $5^n$ is congruent in turn to $1$, $5$, $7$, $8$, $4$, and $2$ modulo $9$. So the pattern does indeed continue "backwards."
Remark: For a more general approach, please see Euler's Theorem.
You are looking at numbers of the form $2n$.
Looking at your enumeration, the most obvious thing is that every second number is divisible by $2$ only once. That this happens is clear, as these are the numbers where $n$ from above is odd.
The second pattern you note is that every $3+4k$th term is divisible by $2$ only twice, and these are the only numbers divisible by $2$ twice. In order to make this more clear, remove all the numbers of the form $2(2k+1)$ from the list to get
$$0,2,3,2,4,2,...$$
These $2$'s correspond to numbers of the form $2(2 (2k+1))=8k+4$, which gives you every number divisible by $2$ exactly twice.
If we strip these two obvious patterns, what you write is
$$3,4,3,5,3,4,3,6,3,4,3,4$$
and here there is an error, the last term should not be a $4$ but at $5$ (it corresponds to $96=2^5\cdot 3^1$).
So we actually find the same pattern as in the other two instances, just that everything got a $+2$.
What happens is this: You are removing from the sequence all odd terms and then you recover the same sequence but with a $+1$. Thus the pattern you see is an artefact of the odd numbers having no $2$ divisors and the rule "remove every second term and you receive the same list but with $+1$ to everything".
Best Answer
What you have discovered is the special case $\rm\ a = 2,\ m = n-1\ $ of the following simple identity
$\rm\displaystyle\ a^{m}\:\frac{a^n-1}{a-1}\: =\: \frac{a^{m+n}-1}{a-1} - \frac{a^m-1}{a-1}\: =\: 1+a+\cdots+a^{m+n-1} - (1+a+\cdots+a^{m-1})$
For example, it yields $\ 11111000 = 11111111 - 111\ $ for $\rm\ a = 10,\ m = 3,\ n = 5\:$.
However, except in the special case when the left hand side is an even perfect number, the right hand side is not the (rearranged) sum of the divisors of the left hand side, so it bears little relation to perfect numbers.