Let β be a curve on M. Show that β is a line of curvature on M iff the surface defined by y(u,v)= β(u)+vU(u) is developable. Here U(u) denotes the unit normal to M along β
Suppose $\displaystyle \beta$ is the curvature line, denote $\displaystyle T = \beta_u$ the tangent vector of $\displaystyle \beta$, and let B be the unit tangent vector of M
that is orthogonal to T. Thus T, B is the two principal directions forming a basis of the tangent plane, and denote K, k the corresponding principal curvatures.
In this basis the second fundamental form of M is a diagonal matrix $\displaystyle \mathrm{I\!I}=diag( K, k)$
For the surface $\displaystyle \gamma(u,v)= \beta(u)+v \cdot U(u)$, the two coordinate tangent vectors are:
$\displaystyle \gamma_u = \beta_u + v \cdot U_u$ and $\displaystyle \gamma_v = U$
And since $\displaystyle U_u \cdot U = \frac{1}{2}(U\cdot U)_u=0 $
$\displaystyle U_u\cdot T=(U\cdot T)_u - U\cdot T_u = - U\cdot T_u=-\beta_{uu} \cdot U = -\mathrm{I\!I}_{11}=-K$
$\displaystyle U_u\cdot B=(U\cdot B)_u - U\cdot B_u = - U\cdot B_u = -\mathrm{I\!I}_{12} = 0$
From the above 3 equations we get $\displaystyle U_u=-K T$ (this can also be obtained by the Frenet Serret formulas)
So $\displaystyle \gamma_u = \beta_u + v \cdot U_u = (1-vK)T$ and $\displaystyle \gamma_v = U$
Then the normal vector of this surface is orthogonal to both T and U thus is constant along the v-lines on the surface, which implies the differential of the Gauss map is degenerate along U, thus having a 0 Gaussian curvature and hence developable.
I'll try to show the converse in a new post.
For the converse, since we have
$\displaystyle \gamma_u = T + v U_u$, and $\displaystyle \gamma_v = U$, we can compute the first fundamental form
let $\displaystyle k=\langle T, U_u \rangle$ and $\displaystyle b=\langle U_u, B \rangle$ so that $\displaystyle U_u = kT+bB$
So $\displaystyle \gamma_u = T + v (kT+bB) = (1+vk)T+bB$
$\displaystyle E = (1 + vk)^2 + b^2$, $\displaystyle F = \langle T, U \rangle + \langle v U_u, U \rangle = v\langle U_u, U \rangle = \frac{v}{2} \langle U, U \rangle_u = 0$,
$\displaystyle G = \langle U, U \rangle = 1$
According to Gaussian curvature
Use the formula $\displaystyle K = -\frac{1}{2\sqrt{EG}}\left(\frac{\partial}{\partial u}\frac{G_u}{\sqrt{EG}} + \frac{\partial}{\partial v}\frac{E_v}{\sqrt{EG}}\right)$.
We get $\displaystyle K = -\frac{1}{2\sqrt{E}}\frac{\partial}{\partial v}\frac{E_v}{\sqrt{E}}$
= $\displaystyle -\frac{1}{\sqrt{E}}\frac{\partial}{\partial v}\frac{(1+vk)k}{\sqrt{(1 + vk)^2 + b^2}}$
Only when b = 0 the last term vanishes and then we get $\displaystyle U_u = kT$, according to the post line of curvature question T is a principal vector.
Somebody asked me via private messages that "why the second fundamental form is $\displaystyle \mathrm{I\!I}=diag(K,k)$?", and let me explain here:
The vector $\displaystyle T=\beta_u$ is a unit principal vector, the unit vector $\displaystyle B=U \times T$ is a tangent vector to M and orthogonal to T. So (T, B) form a orthonormal basis of the tangent plane of M. Since T is a principal direction B is the other principal direction, that is, they're the eigen-vectors of the second fundamental form. Then the result follows from the fact that a matrix is diagonal when expressed with respect to its eigen-vector basis, with eign-values as the diagonal values.