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.