It's pretty much a classical result that the catenoid and helicoid are isometric; if you look up a proof of that, you'll probably be able to discover the inconsistency in your reasoning.
One thought: there's no reason that the isometry has to take $x$ to $x$ and $y$ to $y$ in your chosen parameterizations, so your reasoning about the coefficients of dx and dy seems invalid. Although as I recall, in the standard catenoid-to-helicoid isometry, the radial coordinate on the helicoid ($x$) corresponds to the axial coordinate ($x$) on the helicoid, so maybe you're OK.
BTW, the function, for the catenoid, is $e^x + e^{-x}$; you may need some constants in there to make things work out.
BTW 2: there's an animation of the helicoid to catenoid isometry on the Wikipedia page for catenoids
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The problem is that an isometry is a map between surfaces that preserves fundamental forms. If you reparameterize the first surface, you have to reparameterize the second "in the same way" to still preserve the forms. But before you can do that, you need to find SOME isometry between SOME pair of parameterizations, and just picking two and hoping doesn't usually work. :(
Following the Wikipedia page, I find that two workable parameterizations are
\begin{align}
S_1(u, v) &= (\sinh v \,\sin u , \sinh v \,\cos u , u) \\
S_2(u, v) &= (\cosh v \,\cos u , \cosh v \,\sin u , v) \\
\end{align}
In your case, in the first parameterization, you have $x = \sinh v$, and in the second, $f(x) = \cosh v$. Since $\cosh^2 x = 1 + \sinh^2 x$, you have $f(x) = \sqrt{1 + x^2}$, just as you conjectured.
But the actual isometry from $S_2$ to $S_1$ is the map $(x, y) \mapsto (\sinh x, y)$.
So you need to compare the 1st fundamental form for $S_2$ at $(x, y)$ to the first fundamental form for $S_1$ at the point $(\sinh x, y)$. If you try that, things might work out.
You cannot cover a circle with one single parametrization, but you are right, you have to cover the circles which are omitted by your parametrization. By using the same parametrization as before on $c< u < 2\pi+c$ and $ c < v < 2\pi + c$ and on
$d< u < 2\pi+d$ and $ d < v < 2\pi + d$ for some small $c \neq d$ you will acchieve this. The first one, with $c>0$ will cover the circles you omitted up to now with the exception of two points, and the last one will also catch those two points. Of course, even smaller patches would do.
(This does, admittedly, not prove that you cannot do better)
Best Answer
There is an atlas for the $2$-torus $T$ with three charts. If we view $T$ as the result of doing identifications as usual on a square $Q$, then the three open sets are
the complement of the image in $T$ of the horizontal sides of $Q$,
the complement of the imagfe in $T$ of the vertical sides of $Q$, and
a small disk in $T$ centered at the image of the vertices of $Q$.
These are two annuli and a disk.
I am pretty sure you cannot do this with with two charts. If one adds the requirement that the domains of the charts are contractible, this follows from the theory of Lusternik–Schnirelmann category and its lower bound in terms of cup length. But of course, two of my open sets above are not contractible...