The question is interesting and made me go into some detail.
Voss-Weyl Formula
This thread explains well
the caveats one has to keep in mind when using the Voss-Weyl formula for the divergence of a vector field $V$ of any dimension:
$$\tag{1}
\operatorname{div}_gV=\frac{\partial_\mu(\sqrt{\operatorname{det}g}\,V^\mu)}{\sqrt{\operatorname{det}g}}\,
$$
where $g_{\mu\nu}$ is a metric tensor.
In short: The Voss-Weyl formula assumes that $V$ is expressed in a coordinate basis, that is,
$$\tag{2}
V=V^\mu\,\partial_\mu\,.
$$
Cylindrical Coordinates
In cylindrical coordinates $(r,\varphi,z)$ one usually works with a different basis, namely,
$$\tag{3}
\partial_r\,,\;\textstyle\color{red}{\frac{1}{r}}\partial_\varphi\,,\;\partial_z
$$
which is orthonormal w.r.t. the well-known cylindrical metric
$$\tag{4}
g=\begin{pmatrix}1&0&0\\0&r^2&0\\0&0&1 \end{pmatrix}\,.
$$
In the orthonormal basis the vector field $V$ from (2) is
$$\tag{5}
V=V^r_{on}\,\partial_r+\textstyle\frac{1}{r}V^\varphi_{on}\,\partial_\varphi+V^z_{on}\partial_z\,.
$$
The most notable difference is $V^\varphi=\frac{1}{r}V^\varphi_{on}\,.$
Since $\sqrt{\operatorname{det}g}=r$ we find from (1)
\begin{align}
\operatorname{div}_gV&=\frac{\partial_r(rV^r)+\partial_\varphi (r V^\varphi)+\partial_z (rV^z)}{r}\\
&=\frac{\partial(rV^r)}{r}+\partial_\varphi V^\varphi+\partial_z V^z\,.\tag{6}
\end{align}
To get the well-known divergence formula we have to switch to the orthonormal basis:
\begin{align}
\operatorname{div}_gV
&=\frac{\partial_r(r V^r_{on})}{r}+\frac{1}{r}\partial_\varphi V^\varphi_{on}+\partial_z V^z_{on}\\
&=\partial_rV^r_{on}+\frac{V^r_{on}}{r}+\frac{1}{r}\partial_\varphi V^\varphi_{on}+\partial_z V^z_{on}\,.\tag{7}
\end{align}
Hyperbolic Plane
It sounds like, by hyperbolic plane you are referring to the Poincare half-plane model
$$\tag{8}
\mathbb H=\{x,y\in\mathbb R:y>0\}\,,\quad ds^2=\frac{dx^2+dy^2}{y^2}\,.
$$
Its metric tensor is
$$\tag{9}
g=\begin{pmatrix}y^{-2}&0\\0&y^{-2}\end{pmatrix}\,.
$$
Since, $\sqrt{\operatorname{det}g}=y^{-2}$
the divergence of a vector field
$$\tag{10}
V=V^x\partial_x+V^y\partial_y
$$
in the coordinate basis is
\begin{align}
\operatorname{div}_gV&=\frac{\partial_x\,(y^{-2}\,V^x)+\partial_y(y^{-2} V^y)}{y^{-2}}=\frac{y^{-2}\,\partial_xV^x-2y^{-3}\,V^y+y^{-2}\partial_y\,V^y}{y^{-2}}\\
&=\partial_x\,V^x+\partial_y\,V^y-\frac{2V^y}{y}\,,\tag{11}
\end{align}
which confirms your calculation.
It looks tempting to seek for a relationship between (11) and (7) when $V$ does not depend on $\varphi$ as $r$ looks very much like $y$ and $x$ looks very much like $z\,.$
However note that the Euclidean space is flat and the half plane $\mathbb H$ is not. Also: I have highlighted the difference of the bases we express the vectorfields in: cylindrical uses orthonormal basis, $\mathbb H$ uses coordinate basis.
