The normal curvature can be defined in terms of the second fundamental form as follows: the normal curvature $k_n$ of a vector $v \in T_{p}S$ is defined as $k_n = \text{II}_{p}(v)=<-dN_{p}(v), v>$. Recall that the differential of the gauss map is a self-adjoint linear map and so there exists an orthonormal basis $\{e_1, e_2\}$ for $T_{p}S$ such that $-dN_{p}(e_1) = k_{1}e_{1}$ and $-dN_{p}(e_2) = k_{2}e_{2}$.
Thus for an arbitrary direction vector described by your $\varphi$, call it $v$ can we written as a linear combination of some orthonormal basis $\{e_1, e_2\}$ with the above properties as $v = e_{1}\cos\varphi + e_{2}\sin\varphi$.
Now following the definition we get:
\begin{align}
k_{n} & = \text{II}_{p}(v) = -<dN_{p}(v), v> \\
& =-<dN_{p}(e_{1}\cos\varphi + e_{2}\sin\varphi), e_{1}\cos\varphi + e_{2}\sin\varphi> \\
& =<k_{1}e_{1}\cos\varphi+k_{2}e_{2}\sin\varphi, e_{2}\sin\varphi+ e_{1}\cos\varphi> \\
& =k_{1}\cos^2\varphi + k_{2}\sin^2\varphi \\
\end{align}