Consider the hyperbolic plane with coordinates $(u,v)$ and metric $\begin{bmatrix}u^{-2}&0\\0&u^{-2}\end{bmatrix}$. For an orthonormal frame field $F_1=u\partial_u$ and $F_2=u\partial_v$, I found the connection form to be $\omega^1_2=-\frac1udv$ using the Cartan structural equations: $$d\theta^i=\omega^i_j\wedge\theta^j,$$ where $(\theta^i)_i$ is the coframe to $(F_i)^i$ such that $\theta^iF_j=\delta^i_j$.
Let $\alpha=(\alpha_1,\alpha_2)$ be a unit speed geodesic.
Given a vector field $W=w^ie_i$ and a frame field $(F_i)^i$, we can define intrinsically to the surface the covariant derivative along $\alpha$ of $W$: \begin{align} \nabla_{\alpha'}W=\left(w^1\,\!'-w^2\omega^1_2(\alpha')\right)F_1+(w^2\,\!'+w^1\omega^1_2(\alpha'))F_2, \end{align} where I used skew-symmetry for the minus sign in the first term.
But, I don't know what it means to take $\omega^1_2(\alpha')$. I can't directly substitute $\omega^1_2$ which is found with the structural equations with respect to the coframe, because for $w^i=\alpha_i'$, we have
\begin{align*} \nabla_{\alpha'}\alpha'=0&=\left(\alpha_1''+\frac{\alpha_2'}{u}dv\right)F_1+\left(\alpha_2''-\frac{\alpha_1'}{u}dv\right)F_2 \end{align*} which to me doesn't make any sense because of the exterior derivative of $v$ sitting in there.
So, what is $\omega^1_2(\alpha')$?
By the way, this is a homework question - please don't give me the answer directly. I am trying to show $\displaystyle \alpha_1'' -\frac1u (\alpha_1')^2 +\frac1u (\alpha_2')^2 = 0$ and $\displaystyle \alpha_2'' -\frac2u \alpha_1' \alpha_2' = 0$, and I can't for the life of me find how to show this with my expression of $\nabla_{\alpha'}\alpha'$.
I've had my question answered - I just forgot basic dual/vector multiplication. For a 1-form $\omega=a\,dx+b\,dy$ and a vector $v=c\partial_x+d\partial_y$, $$\omega(v)=ac+bd.$$ That's it. I was overcomplicating it.