Given $p/q, r/s \in \mathbb{Q}$, where $\gcd(p,q)=\gcd(r,s)=1$, choose the triangles in the Farey tessellation whose interior intersects the geodesic connecting $p/q$ and $r/s$ in $\mathbb{H}^2$:

If $ps-qr=1$, then $d(p/q,r/s)=1$ and there is a geodesic of the Farey tessellation
connecting the two points. Otherwise, there will be a non-trivial triangle intersecting the geodesic. The triangles may be grouped together into maximal collections of triangles sharing a common vertex (a "pivot"). Let $p_i$ be the number of triangles in the $i$th group.
Associated to this sequence of grouped triangles is a canonical sequence $[p_1,p_2, \ldots, p_k]$,
where $p_1>1, p_k>1$. Define $l([p_1,p_2,\ldots,p_k])=d(p/q,r/s)$.

Note that the sequence of triangles $[1,p_2, \ldots, p_k]$ is the same as the
sequence $[1+p_2, \ldots, p_k]$, since the lone triangle shares a vertex with the
pivot of the adjacent group, which is why we may assume that $p_1>1$,
and similarly $p_k>1$.

The formula for the the Farey distance is computed inductively by
$l([p_1,p_2,\ldots,p_k]) = 1+ l([p_2,\ldots,p_k])$, and for $k=1$,
$l([p_1])= 2$ (we are assuming that $p_1>1$ when $k=1$, since otherwise $d(p/q,r/s)=1$).
Here, if $p_2=1$, then $l([p_2,\ldots,p_k])=l([1+p_3,\ldots,p_k])$.
To prove this formula, one
can see that a shortest path from $p/q$ to $r/s$ must go through
the adjacent pivot to $p/q$. If not, a case-by-case analysis shows that one
can find a shorter path, unless $p_1=2$ and the path goes through
the lower two edges of the first group of triangles. But then one can take an equal length path going
through the first pivot.

To relate this to the continued fraction expansion, use an element of $A\in PGL_2(\mathbb{Z})$ such that $A(p/q)=\infty$, and $0\leq A(r/s) \leq 1/2$. Then $d(p/q,r/s)=d(A(p/q),A(r/s))=d(\infty,A(r/s))$. To compute the triangle
sequence in this case, we may use the continued fraction expansion for $A(r/s)$ to get $[p_1,\ldots,p_k]$ via $A(r/s)=1/(p_1+1/(p_2+1/(\cdots +1/p_k)\cdots )))$:

In this example, $p/q=\infty=1/0$, $r/s=2/5$, and the continued fraction is $[2,2]$, since $2/5=1/(2+1/2)$.