I thought about your problem and realized that there is no coordinate parametrization of the catenoid with the properties that you want.
Here is my argument: First, note that, in the given $uv$-parametrization, the first and second fundamental forms of the immersion are
$$
I= \cosh(u)^2 (\mathrm{d}u^2 + \mathrm{d}v^2)
\quad\text{and}\quad
I\!I = \mathrm{d}u^2- \mathrm{d}v^2.
$$
We want to find two (possibly local) foliations by geodesics,
say $\mathcal{F}_i$ for $i=1,2$, such that the tangent vectors to the leaves of one foliation are conjugate to the tangent vectors to the leaves of the other foliation. What this means is that,
if $X$ is a vector field tangent to the leaves of $\mathcal{F}_1$ and $Y$ is a vecor field tangent to the leaves of $\mathcal{F}_2$, then $I\!I(X,Y)=0$.
Of course, we can assume, without loss of generality, that $X$ and $Y$ are unit vector fields, i.e., $I(X,X)=I(Y,Y)=1$.
Now, given any unit tangent vector field to the surface, say, $Z$, the condition that the flow lines of $Z$ be geodesics is that $Z^\flat$, the dual $1$-form, be closed.
Thus, we require that $X$ and $Y$ be unit vector fields so that $I(X,X)=I(Y,Y)=1$ and that $X^\flat$ and $Y^\flat$ be closed, and we also require that $I\!I(X,Y) = 0$, since this is the conjugate property. If
$$
X^\flat = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v),
$$
for some angle function $\alpha$, then
the equations $I(Y,Y)=1$ and $I\!I(X,Y) = 0$ force
$$
\pm Y^\flat = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v).
$$
Note that $\cos(2\alpha)\not=0$, otherwise, $X\pm Y = 0$, and the two foliations will not be transverse, so they can't be the level sets of the desired coordinate functions.
So we need that the following $1$-forms be closed:
$$
\xi = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v) \quad\text{and}\quad
\eta = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v).
$$
If we take the exterior derivatives of these 1-forms and set them equal to zero, we find that $\alpha$ must satisfy the equation
$$
\mathrm{d}\alpha = \frac{\tanh u}{\cos 2\alpha}\,
\bigl(\mathrm{d}v - \sin 2\alpha\, \mathrm{d}u\bigr).
$$
But now applying the exterior derivative to both sides of this relation, we see that
$$
0 = \frac{1-2\cosh^2u}{\cosh^2u\,\cos 2\alpha}\,\mathrm{d}u\wedge\mathrm{d}v.
$$
Since $1-2\cosh^2u$ is never zero, we have a contradiction, so $\alpha$ cannot exist.
Remark: For comparison, it might be helpful to look at the case of the helicoid, parametrized by $(u\cos v, u\sin v, v)$. Here we have
$$
I= \mathrm{d}u^2 + (u^2{+}1)\mathrm{d}v^2
\quad\text{and}\quad
I\!I = \frac{2 du\,dv}{\sqrt{u^2{+}1}}.
$$
now, reasoning as above, we seek an angle $\alpha$ such that the $1$-forms
$$
\xi = \cos\alpha\,\mathrm{d}u +\sin\alpha\,\sqrt{u^2{+}1}\,\mathrm{d}v
\quad\text{and}\quad
\eta = \cos\alpha\,\mathrm{d}u -\sin\alpha\,\sqrt{u^2{+}1}\,\mathrm{d}v
$$
are closed. The equation $\mathrm{d}(\xi+\eta)=2\,\mathrm{d}(\cos\alpha\,\mathrm{d}u)=0$ implies that $\alpha$ must
be a function of $u$ and the equation $\mathrm{d}(\xi-\eta)=2\,\mathrm{d}(\sin\alpha\,\sqrt{u^2{+}1}\,\mathrm{d}v)=0$ then implies
that $\sin\alpha\,\sqrt{u^2{+}1} = c$ for some constant $c\not=0$.
Thus,
$$
\xi = \sqrt{\frac{u^2{+}1{-}c^2}{u^2{+}1}}\,\mathrm{d}u +c\,\mathrm{d}v
\quad\text{and}\quad
\eta = \sqrt{\frac{u^2{+}1{-}c^2}{u^2{+}1}}\,\mathrm{d}u -c\,\mathrm{d}v,
$$
defined on the range where $u^2{+}1{-}c^2>0$ (which is everywhere if $|c|<1$). The dual vector fields are
$$
X_\pm = \sqrt{\frac{u^2{+}1{-}c^2}{u^2{+}1}}\,
\frac{\partial}{\partial u} \pm \frac{c}{u^2{+}1}\,\frac{\partial}{\partial v},
$$
so we choose $x$ and $y$ so that $\mathrm{d}x$ and $\mathrm{d}y$
each annihilate one of $X_\pm$, say
$$
\mathrm{d}x = \mathrm{d}v
+\frac{c\,\mathrm{d}u}{\sqrt{(u^2{+}1{-}c^2)(u^2{+}1)}}
\quad\text{and}\quad
\mathrm{d}y = \mathrm{d}v
-\frac{c\,\mathrm{d}u}{\sqrt{(u^2{+}1{-}c^2)(u^2{+}1)}},
$$
which gives a 1-parameter family of solutions to the problem.
