Yes, for given $ u(t), v(t)$ the $\alpha(t), \beta (t) $ are both lines of principal curvature since derivatives vanish, and normal curvatures are maximum and minimum for these curvature lines.
For this to happen diagonal coefficients as also next mixed coefficients $ F, (f = M) $ of first and second fundamental form coefficients vanish.
Incidentally, other non-principal direction angular deviation on surface $ \psi $ can be referenced through $ u', v'$ with Euler relation for principal normal curvatures as $ k_n = k_1 \cos^2\psi + k_2 \sin^2 \psi. $
EDIT1:
Regarding Y(t,0)=α(t) and Y(0,t)=β(t) separation of functions on a single independent variable t.
While referring to a point on a surface we can never get rid of one of the parameters, but we can make it constant. It is like forgetting the road we came by after turning at an intersection.
We say Y(u,v) for any point referred on the surface, $Y(u,v1)$ for $v = v1$ = constant or $Y(u1,v)$ for $u = u1$ = another constant, Y(u,0) for the v = 0 line, Y(0,v) for the u = 0 line. The parametrization Y(u,v) implies we are referring to particular value of parameter u= u_1 = constant or v = v_1 = constant.
The following text-book material of surface theory is relevant to go through:
($E,F,G$ first fundamental form metric coefficients, $ e,f,g = L,M,N $ second fundamental form metric coefficients). The normal curvature
$$ k_n =\dfrac {L du^2 + 2 M du dv + N dv^2 }{E du^2 + 2 F du dv + G dv^2} $$
$$ = L (du/ds)^2 + 2 M (du/ds) (dv/ds) + N (dv/ds)^2 $$
Line of curvatures
$$ k_n = {(E M-F L) du^2 + (E N-G L) du dv + (F N-G M) dv^2 } $$
Necessary and sufficient condition for a line of curvature to be parametric $$ F =0, M =0 $$
Rodrigue's formula necessary and sufficient relation:
$$ k_n = - dN(u,v)/dY(u,v) $$
To derive Euler $k_n$ formula because $F=0$ ,
$$ \cos\psi = \sqrt {E} (du/ds) \;,\sin\psi =\sqrt {G} (dv/ds) $$
For a sphere lines of curvature ( meridians $\theta$ = const and parallels $\phi $ = constant ) are:
$$ x = a \cos\phi \cos \theta \;; y= a \cos\phi \sin\theta \; ; z = a \sin\phi ; $$
along a meridian $ \phi= \alpha $
Best Answer
Let $X\colon U\subset \mathbb{R}^2\longrightarrow S\subset \mathbb{R}^3$ be a parametrization of a regular surface $S$ that contains no umbilical points of $S$. Then we will show that:
the parametrized curves of $X$ are lines of curvature $\Longleftrightarrow$ $F=f=0$.
Proof: