I think I've done everything correct, but I'd like to have a second pair of eyes to look at it. So enjoy the calculations!
Consider the curved cylinder with the parametrisation
$$F:\mathbb{R}\times[0,2\pi)\rightarrow\mathbb{R}^3,\quad F(t,\phi)=\big((1+t^2)\cos\phi,(1+t^2)\sin\phi,t\big)$$
It basically looks like the left of those cylinders
I calculate the tangent and normal vectors:
$$\begin{align} \frac{dF}{dt}&= (2t\cos\phi,2t\sin\phi,1)\\ \frac{dF}{d\phi}&=(1+t^2)(-\sin\phi,\cos\phi,0)\\ \frac{d^2F}{dt^2}&=2(\cos\phi,\sin\phi,0)\\ \frac{d^2F}{dtd\phi}&=2t(-\sin\phi,\cos\phi,0)\\ \frac{d^2F}{d\phi^2}&=-(1+t^2)(\cos\phi,\sin\phi,0)\\ \frac{dF}{dt}\times\frac{dF}{d\phi}&=-(1+t^2)(\cos\phi,\sin\phi,-2t)\\ \|\frac{dF}{dt}\times\frac{dF}{d\phi}\|&=(1+t^2)\sqrt{1+4t^2}\\ N\circ F&=-\frac{1}{\sqrt{1+4t^2}}(\cos\phi,\sin\phi,-2t) \end{align} $$
For the first and second fundamental form and the curvatures I find
$$ \begin{align} g&=\begin{pmatrix} 1+4t^2 & 0\\ 0 & (1+t^2)^2 \end{pmatrix}\\ h&=\begin{pmatrix} -\frac{2}{\sqrt{1+4t^2}} & 0\\ 0 & \frac{1+t^2}{\sqrt{1+4t^2}} \end{pmatrix}\\ K&=\frac{\det(h)}{\det(g)}=-\frac{2}{(1+4t^2)(1+t^2)}\\ H&=\frac{1}{2}\operatorname{trace}(hg^{-1})=\frac{1}{2}\cdot\frac{2t^2-1}{(1+4t^2)^\frac{3}{2}(1+t^2)} \end{align} $$
And for the Christoffel-symbols
$$ \begin{align} \Gamma^1_{11}&=\frac{4t}{1+4t^2}\\ \Gamma^1_{12}=\Gamma^1_{21}&=0\\ \Gamma^1_{22}&=-\frac{2t(1+t^2)}{1+4t^2}\\ \Gamma^2_{11}&=0\\ \Gamma^2_{12}=\Gamma^2_{21}&=\frac{2t}{1+t^2}\\ \Gamma^2_{22}&=0\\ \end{align} $$
The covariant vectors are
$$ \begin{align} \nabla_{\frac{dF}{dt}}\frac{dF}{dt}&=\Gamma^1_{11}\frac{dF}{dt}+\Gamma^2_{11}\frac{dF}{d\phi}=\frac{4t}{1+4t^2}\frac{dF}{dt}\\ \nabla_{\frac{dF}{dt}}\frac{dF}{d\phi}=\nabla_{\frac{dF}{d\phi}}\frac{dF}{dt}&=\Gamma^1_{12}\frac{dF}{dt}+\Gamma^2_{12}\frac{dF}{d\phi}=\frac{2t}{1+t^2}\frac{dF}{d\phi}\\ \nabla_{\frac{dF}{d\phi}}\frac{dF}{d\phi}&=\Gamma^1_{22}\frac{dF}{dt}+\Gamma^2_{22}\frac{dF}{d\phi}=-\frac{2t(1+t^2)}{1+4t^2}\frac{dF}{dt}\\ \end{align}$$
Question: Can I use those covariant vectors to calculate a geodesic?
Now I check if the meridian and the latitudes are geodesics
$$c(t)=F(t,\phi)=F(\tilde{c}(t))$$ with $$\tilde{c}_1(t)=t,\quad \tilde{c}_2(t)=\phi$$ The derivates are $$\tilde{c}_1'(t)=1,\quad \tilde{c}_2'(t)=0$$ $$\tilde{c}_1''(t)=0,\quad \tilde{c}_2''(t)=0$$ Therefore the geodesics equations
$$\tilde{c}_k''(t)+\Gamma^k_{ij}(\tilde{c}(t))\tilde{c}_i'(t)\tilde{c}_j'(t)$$ simplify to
$$ \begin{align} \Gamma^1_{11}(t,\phi)&=\frac{4t}{1+4t^2}\ne 0\\ \Gamma^2_{11}(t,\phi)&=0 \end{align}$$
Therefore the meridians are no geodesics
Now I do the same for the latitudes:
$$c(\phi)=F(t,\phi)=F(\tilde{c}(\phi))$$ with $$\tilde{c}_1(\phi)=t,\quad \tilde{c}_2(\phi)=\phi$$ The derivates are $$\tilde{c}_1'(\phi)=0,\quad \tilde{c}_2'(\phi)=1$$ $$\tilde{c}_1''(\phi)=0,\quad \tilde{c}_2''(\phi)=0$$ Therefore the geodesics equations
$$\tilde{c}_k''(\phi)+\Gamma^k_{ij}(\tilde{c}(\phi))\tilde{c}_i'(\phi)\tilde{c}_j'(\phi)$$ simplify to
$$ \begin{align} \Gamma^1_{22}(t,\phi)&=-\frac{2t(1+t^2)}{1+4t^2}\\ \Gamma^2_{22}(t,\phi)&=0 \end{align}$$
For the latitude at $t=0$, I find that both equations vanish. Therefore the latitude $$c_0(\phi):=F(0,\phi)$$ is a geodesic.
Finally I want to calculate the geodesic curvature of the latitude at $t=1$, i.e. the value
$$\kappa_g=\langle\frac{\nabla}{d\phi}\gamma'(\phi),\eta(\phi)\rangle$$
where $\gamma$ is the same curve as $F(1,\cdot)$, but parametrised by arclength.
Define the curve $$c(\phi)=F(1,\phi)=F(\tilde{c}(\phi))$$
with
$$\tilde{c}_1(\phi)=1,\quad \tilde{c}_2(\phi)=\phi$$
First, the derivates are
$$\tilde{c}_1'(\phi)=0,\quad \tilde{c}_2'(\phi)=1$$
I calculate the first derivative and check if it is parametrised by arclength:
$$c'(\phi)=\frac{d}{ds}F(\tilde{c}(\phi+s))\big|_{s=0}=\frac{dF}{d\phi}(\tilde{c}(\phi))$$
$$\|c'(\phi)\|^2=g_{22}(\tilde{c}(\phi))=(1+t^2)^2\big|_{\tilde{c}(\phi)=(1,\phi)}=4\ne 1$$
Now parametrise this curve by arclength:
$$\alpha^{-1}(\phi):=\int_0^\phi 2ds=2\phi$$ has the inverse function
$$\alpha(\phi)=\frac{\phi}{2}$$
with derivative
$$\alpha'(\phi)=\frac{1}{2}$$
Define the curve $\gamma:=c\circ\alpha$, which is parametrised by arclength. Its derivates are
$$\begin{align} \gamma'(\phi)&=\alpha'(\phi)c'(\alpha(\phi))=\frac{1}{2}\frac{dF}{d\phi}(\tilde{c}(\alpha(\phi)))\\ \gamma''(\phi)&=\frac{1}{2}\frac{d}{ds}\frac{dF}{d\phi}(\tilde{c}\circ\alpha(\phi+s))\big|_{s=0}\\ &=\frac{1}{2}\langle\operatorname{grad}\frac{dF}{d\phi}(\tilde{c}(\alpha(\phi)),\tilde{c}'(\alpha(\phi))\cdot\alpha'(\phi)\rangle\\ &=\frac{1}{4}\frac{d^2F}{d\phi^2}(\tilde{c}(\alpha(\phi))) \end{align}$$
And the covariant derivate is
$$ \begin{align} \frac{\nabla}{d\phi}\gamma'(\phi) &=\frac{1}{4}\Big(\Gamma^1_{22}\frac{dF}{dt}+\Gamma^2_{22}\frac{dF}{d\phi}\Big)(\tilde{c}(\alpha(\phi)))\\ &=\Big(\frac{1}{4}(-\frac{2t(1+t^2)}{1+4t^2})\frac{dF}{dt}\Big)(1,\frac{\phi}{2})\\ &=-\frac{1}{5}\frac{dF}{dt}(\tilde{c}(\alpha(\phi))) \end{align}$$
Since $g_{12}=g_{21}=0$ and $(\frac{dF}{d\phi},-\frac{dF}{dt})$ is apositive oriented base of the tangent space, I choose
$$\eta(\phi)=-\|\frac{dF}{dt}\|^{-1}\cdot\frac{dF}{dt}(\tilde{c}(\alpha(\phi)))=-\Big(\frac{1}{\sqrt{1+4t^2}}\frac{dF}{dt}\Big)(\tilde{c}(\alpha(\phi)))=-\frac{1}{\sqrt{5}}\frac{dF}{dt}(\tilde{c}(\alpha(\phi)))$$
as a united speed normal vector, such that $$(\gamma',\eta)$$ forms a positive oriented orthogonal base of the tangent space.
Finally I calulate
$$ \begin{align} \kappa_g &=\langle\frac{\nabla}{d\phi}\gamma'(\phi),\eta(\phi)\rangle\\ &=\Big(-\frac{1}{5}\Big)\Big(-\frac{1}{\sqrt{5}}\Big)\langle\frac{dF}{dt},\frac{dF}{dt}\rangle(\tilde{c}(\alpha(\phi)))\\ &=\Big(-\frac{1}{5}\Big)\Big(-\frac{1}{\sqrt{5}}\Big)g_{11}(\tilde{c}(\alpha(\phi)))\\ &=\Big(-\frac{1}{5}\Big)\Big(-\frac{1}{\sqrt{5}}\Big)\Big(1+4t^2\Big)\Big|_{\tilde{c}(\alpha(\phi)))=(1,\frac{\Phi}{2})}\\ &=\frac{1}{\sqrt{5}} \end{align} $$
If anyone has been reading until here, I am very grateful for your attention! Feel free to make any reply or comment on my post. Especially if you find, that anything is wrong, I would be very happy, to hear from you. Also if you find that the calculations are right, I'd be happy for a short reply. :-)
Thanks again for your attention!