Question about Normal Curvature

Question

I am trying to show that the normal curvature of M in the direction of α′ is equal to the normal component of acceleration. I know that if I start with $U\cdot \alpha'=0$ and derive that I end up with \begin{align*} U'\cdot \alpha' + U\cdot \alpha'' =0 \end{align*} And then I can subtract $U'\cdot \alpha'$ to the other side. But how is this related to the shape operator? I know $S(\alpha')=U\cdot \alpha'$, not $U'\cdot \alpha'$. My goal is to somehow get to $S(\alpha')\cdot\alpha'$.