Assume that you have a parametrization of a surface $f:\Omega \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3,(u,v) \mapsto f(u,v)$.
Now if I have a curve defined by $g(t)=f(0,t)$.
The geodesic equation(just one component) is given by $$\gamma''_u + 2 \Gamma_{uv}^u \gamma'_u \gamma'_v =0$$
Now I was wondering: How exactly is $\gamma''_u$ for example defined in terms of the curve $g$?
So assume anybody told you that $g$ is a geodesic and you wanted to check this. How could this be done? If anything is unclear, please let me know.
The dashes just denote the derivative of your curve (which is a function of a real variable). I'm going to use dots instead of dashes to stop them from getting in the way of the indices.
Your geodesic equation doesn't seem to be correct - the general expression is
$$ \ddot \gamma^i = \Gamma^i_{jk} \dot\gamma^j \dot\gamma^k,$$
so even if you're taking the $u$ component it should be
$$ \ddot \gamma^u = \Gamma^u_{jk} \dot \gamma^j \dot \gamma^k$$
where $j$ and $k$ are both summed over $\{u,v\}$.
Your particular curve $\gamma(t) = f(0,t)$ has coordinate representation $(\gamma^u,\gamma^v)(t) = (0,t)$. Just differentiating this gives you $\dot \gamma = (0,1) = \partial_v$ and $\ddot \gamma = 0$. Thus the geodesic equation for $\gamma$ is simply $$0=\Gamma^i_{vv}.$$
So to check that this particular $\gamma$ is a geodesic, you need to compute the symbols $\Gamma$ in your particular parametrization and check that the components $\Gamma^u_{vv}, \Gamma^v_{vv}$ are zero.