Here's one approach. Let's say you want to find the beginning of sample $k$.
There are $k-1$ samples ahead of this sample, and hence $10(k-1)$ bits. Thus you want to find the position of the $10(k-1)+1=(10k-9)$th bit.
Write $10k-9=8q+r$ where $q=\lfloor \frac{10k-9}{8}\rfloor$ and $r=(10k-9)\% 8$ (i.e., quotient and remainder in dividing $10k-9$ by $8$).
Then $q$ will give the row number, and $r-1$ will give the bit number. (I think this takes into account that you start your addressing with $0$.) Note that $r$ is always odd, so $r-1$ never puts you in the previous row.
Example: Find (the beginning of) sample $7$.
$10\cdot 7-9=61=8\cdot 7+5$. Sample $7$ should start in row $7$, bit $4$. (With row and bit numbering starting at $0$ as in the helpful chart you provided with the question!)
There is a direct analogous when you work with base $10$.
Take the number $3$ in base $10$. Shift it left: you get $30$, which is $3 \cdot 10$ (and the factor $10$ appears because you are working with base $10$).
The same applies to base $2$. Shifting left is the same as multiplying by $2$.
This comes from the use of positional notation to denote numbers (https://en.wikipedia.org/wiki/Positional_notation).
In base $b$ ($b>1$) the second digit from the right counts $b$ times more than the first digit from the right, the third from the right counts $b$ times more than the second from the right (or $b^2$ times more than the first from the right), and so on.
When you write a number like this
$$ a_n a_{n-1} \dots a_2 a_1 a_0 $$
(in base $b$), what you actually mean is the following
$$ a_n \cdot b^n + a_{n-1} \cdot b^{n-1} + \dots + a_2 \cdot b^2 + a_1 \cdot b^1 + a_0 \cdot b^0.$$
With this in mind one can show that the two operations of shifting left and multiplying by $b$ are actually the same:
$$ a_n a_{n-1} \dots a_2 a_1 a_0 0 = \\ = a_n \cdot b^{n+1} + a_{n-1} \cdot b^{n} + \dots + a_2 \cdot b^3 + a_1 \cdot b^2 + a_0 \cdot b^1 + 0 \cdot b^0 =\\= b \cdot \left(a_n \cdot b^n + a_{n-1} \cdot b^{n-1} + \dots + a_2 \cdot b^2 + a_1 \cdot b^1 + a_0 \cdot b^0 \right) =\\= b \cdot (a_n a_{n-1} \dots a_2 a_1 a_0). $$
Best Answer
If we start counting from $0$ the bits in even position are worth $1,4,16,\ldots 2^{2k}$. When you divide any of these by $3$ you leave a remainder $1$. The bits in odd position are worth $2,8,32,\ldots 2^{2k+1}$. When you divide any of these by $3$ you leave a remainder of $2$ or, equivalently, $-1$. The remainder of dividing by $3$ is then the number of bits in even position minus the number of bits in odd position.