Short answer: the curvature is a property of the second derviative, and it's perfectly possible for a $C^1$ function to have a badly behaved second derivative.
In one variable, think about the function $f(t) = |t|^{3/2}$; it is $C^1$ but $f''(0) = +\infty$.
To turn this into a surface with infinite curvature, try
$$X(u,v) = (u, |u|^{3/2}, v)$$
i.e. just extending this curve in the $z$ direction, forming a sheet. You'll find the normal curvature at the origin in the direction normal to the $xy$-plane is infinite.
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
First of all, your examples of $X_u$ and $X_v$ can't occur (note that $X_{uv}\ne X_{vu}$).
You need more information than just the second partial derivatives to get the principal directions and curvatures. You need to calculate the second fundamental form matrix $$\mathbf{II} = \begin{bmatrix} X_{uu}\cdot n & X_{uv}\cdot n \\ X_{vu}\cdot n & X_{vv}\cdot n \end{bmatrix},$$ where $n$ is the unit normal vector, the first fundamental form matrix $$\mathbf{I} = \begin{bmatrix} X_u\cdot X_u & X_u\cdot X_v \\ X_v\cdot X_u & X_v\cdot X_v\end{bmatrix},$$ and then calculate the matrix of the shape operator by taking $\mathbf I^{-1}\mathbf{II}$. This is the matrix whose eigenvalues and eigenvectors you want to find.
By the way, you may find my freely downloadable differential geometry text (see my profile) of use.