Going through the following lecture notes from MIT on differential geometry - Link
At equation 3.25, in order to calculate the magnitude of the normal curvature, an expression obtained from differentiating $N\cdot t = 0$ along a curve with respect to arclength $s$ was used.
I understand what happened until this point, however, the jump from Eq.3.25 to Eq.3.26 is a little fuzzy to me.
Also, the transition from Eq.3.27 to Eq.3.28 is not entirely clear as how the Normal $N$ is not being differentiated anymore, and the surface $r$ is being differentiated twice now instead of once (from $r_u$ to $r_uu$).
I would appreciate any tips or directions that may help in making things clearer.
From eq. 3.25 to eq. 3.26: Note that $$ \begin{align*} d\mathbf{r} &= \mathbf{r}_u du +\mathbf{r}_v dv,\\ d\mathbf{N} &= \mathbf{N}_u du + \mathbf{N}_v dv. \end{align*} $$ First you have to expand the inner products $d\mathbf{r}\cdot d\mathbf{N}$ and $d\mathbf{r}\cdot d\mathbf{r}$, then you have to replace the expressions for $L$, $M$, $N$ and $E$, $F$, $G$.
From eq. 3.27 to eq. 3.28: Note that $\mathbf{r}_u \cdot \mathbf{N}=0$. Therefore we have $$ \begin{align*} 0 = (\mathbf{r}_u \cdot \mathbf{N})_u = \mathbf{r}_{uu}\cdot \mathbf{N} + \mathbf{r}_u \cdot \mathbf{N}_u, \end{align*} $$ so $L= \mathbf{r}_{uu}\cdot \mathbf{N}$. The alternative expressions for $M$ and $N$ can be found in a similar way.