How do I directly find the $n$th multiplication by $2$ of a number, $x$?
$$ f(x) \implies S_x \mid x_n \in S_x \implies n = 1 \lor x_n = x_{n-1}\times 2 \mid n \in \mathbb{N} $$
Examples:
$$ f(1) = 2^{n-1} \implies S_1 = \{1, 2, 4, 8, ...\} \\ f(2) = 2^n \implies S_2 = \{2, 4, 8, 16, ...\} \\ f(3) = 2^n + 2^{n-1} \implies S_3 = \{3, 6, 12, 24, ...\} \\ f(4) = 2^n + 2^n \implies S_4 = \{4, 8, 16, 32, ...\} \\ f(5) = 2^{n+1} + 2^{n-1} \implies S_5 = \{5, 10, 20, 40, ...\} \\ f(6) = 2^{n+1} + 2^n \implies S_6 = \{6, 12, 24, 48, ...\} \\ f(7) = 2^{n+1} + 2^{n} + 2^{n-1} \implies S_7 = \{7, 14, 28, 56, ...\} \\ f(228517) = 2^{n+16} + 2^{n+15} + 2^{n+13} + 2^{n+12} + 2^{n+11} + 2^{n+10} + 2^{n+9} + 2^{n+6} + 2^{n+4} + 2^{n+1} + 2^{n-1} \implies S_{228517} = \{228517, 457034, 914068, 1828136, ...\} \\ $$
What is the name of the transformation between $x, f(x), S_x$ used in the examples?
What is the relationship between the distance between $2$ and $x$ and $f(x)$?
The repeated multiplication is called a geometric progression.
Finding the nth term: multiply the initial term by the common ratio, to the power of n minus one.
$$ a_n = ar^{n-1} \\ $$
The formulas you were using were power of two combinations representing the initial term. Raising n does multiply by 2, so you get a factor of 2 and the initial term.