$\newcommand{\diff}{\mathrm{d}}$This is a heavily edited version of my (a bit right, but mostly wrong) previous answer.  I think that this is now correct (even got the signs right!). 

----

$\newcommand{\HH}{\mathbb{H}}$Let $L = L_{\alpha, \beta}$ be the semicircle in $\HH$ (the upper-half plane) which is perpendicular to the real axis and which has end points at $\alpha$ and $\beta$.  Breaking the symmetry of the situation, we assume that $\alpha < \beta$.  We find that the centre of $L$ is $c = \frac{1}{2}(\alpha + \beta)$ and the radius of $L$ is $r = \frac{1}{2}(- \alpha + \beta)$.  For later use we take $a = c + ir$ to be the *apex* (highest point) of $L$.

Note that $\diff c = \frac{1}{2}(\diff\alpha + \diff\beta)$ and $\diff r = \frac{1}{2}(-\diff\alpha + \diff\beta)$.  Thus $\diff c\,\diff r = \frac{1}{4} \diff\alpha\, \diff\beta$.  We deduce that: 
$$\frac {\diff c\, \diff r}{r^2} = \frac{1}{4} \frac{\diff\alpha\, \diff\beta}{((\alpha - \beta)/2)^2} = \frac{\diff\alpha\, \diff\beta}{(\alpha - \beta)^2}$$

Suppose now that $v$ is an element of $T_1 \HH$ - that is, a unit tangent vector in the upper half plane.  Suppose that $v$ is tangent to the semicircle $L = L_{\alpha, \beta}$.  Suppose that the basepoint $(x, y)$ of $v$ is (signed) hyperbolic distance $t$ from $a$, the apex of $L$.  Finally, suppose that $v$ makes angle $\theta$ with the horizontal line through $(x, y)$.  It is now an [exercise][1] to show the following:
$$\sin(\theta) = \tanh(t)$$
(Hint: Draw a [picture][2].) Being careful with signs, we compute the following:
$$
\begin{align*}
x      &= c - r \sin(\theta) \\
y      &= r \cos(\theta)
\end{align*}
$$
Thus:
$$
\begin{align*}
\diff x &= \diff c - \sin(\theta)\,\diff r - r \cos(\theta)\,\diff \theta \\
\diff y &= \cos(\theta)\,\diff r - r \sin(\theta)\,\diff \theta \\
\diff \theta &= \cos(\theta)\,\diff t
\end{align*}
$$
We substitute to remove $\diff \theta$ from the right-hand side and obtain:
$$
\begin{align*}
\diff x &= \diff c - \sin(\theta)\,\diff r - r \cos^2(\theta)\,\diff t \\
\diff y &= \cos(\theta)\,\diff r - r \sin(\theta) \cos(\theta)\,\diff t \\
\diff \theta &= \cos(\theta)\,\diff t
\end{align*}
$$
Thus we have: 
$$
\diff x\,\diff y\,\diff \theta = \cos^2(\theta)\,\diff c\,\diff r\,\diff t
$$
Since $y = r \cos(\theta)$, we finally deduce that the two measures given by Series are equal. 

Now, suppose that $\phi_s \colon T_1 \HH \to T_1 \HH$ is the time $s$ geodesic flow.  This takes $v$ to $\phi_s(v)$, the tangent vector with coordinates $(\alpha, \beta, t + s)$.  (That is, $\phi_s(v)$ is again tangent to $L = L_{\alpha, \beta}$ but now has signed distance $t + s$ from the apex.)  Thus $\phi_s^*(\diff\alpha) = \diff\alpha$, $\phi_s^*(\diff\beta) = \diff\beta$, and $\phi_s^*(\diff t) = \diff t$, as desired.


  [1]: https://en.wikipedia.org/wiki/Gudermannian_function
  [2]: https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model#/media/File:Distance_in_the_half-plane_model_3.png