Building on Dennis #19's comment:

Take a row, and split each fraction \\(\frac{p}{q}\\) up into two adjacent fractions in the following row:

$$\frac{p}{q} \rightarrow \frac{p}{p + q},\frac{p + q}{q}$$

This procedure generates the next row.

This gets us a little way, I think, towards understanding the doubling of the terms in each row.

EDIT:

Assume that sequence of iterates corresponds to the infinite tree generated by the root \\(\frac{1}{1}\\) and the production rule \\(\frac{p}{q} \rightarrow \frac{p}{p + q}, \frac{p + q}{q}\\). Then each positive rational number appears once in the tree.

Existence: Fix \\(\frac{p}{q} \in \mathbb{Q}_{+}\\), and we will construct the path back to the root. If \\(\frac{p}{q} = 1\\), then we're done. If \\(\frac{p}{q} > 1\\), then we can write \\(\frac{p}{q} = \frac{p' + q}{q}\\) for some integer \\(p' > 0\\), and \\(\frac{p' + q}{q}\\) is the right descendant of \\(\frac{p'}{q}\\) . Similarly, If \\(\frac{p}{q} < 1\\), then we can write \\(\frac{p}{q} = \frac{p}{p + q'}\\) for some integer \\(q' > 0\\), and \\(\frac{p}{p + q'}\\) is the left descendant of \\(\frac{p}{q'}\\). In either recursive case, we will have either \\(p'\\) strictly less than \\(p\\), or \\(q'\\) strictly less than \\(q\\). Neither \\(p'\\) nor \\(q'\\) can be non-positive, so the process must terminate at the root \\(\frac{1}{1}\\).

Uniqueness: To see that this construction guarantees uniqueness, notice that there is no choice in traversing up the tree. So if \\(\frac{p}{q}\\) occurs at (p)ositions \\(\pi_1\\) and \\(\pi_2\\) in the tree, then we can use the above construction to find p(a)ths \\(A_1\\) and \\(A_2\\) back to the root. But we must have \\(A_1 = A_2\\) by construction, and since the paths are invertible, we must have \\(A_1^{-1} = A_2^{-2}\\) as well, hence \\(\pi_1 = A_1^{-1}(\frac{1}{1}) = A_2^{-2}(\frac{1}{1}) = \pi_2\\). Therefore the position of \\(\frac{p}{q}\\) is unique.

This all hangs, of course, upon some way of connecting the original definition of \\(f\\) to the tree construction.