This is correct. The normal bundle to a divisor in $X$ measures the self-intersection of the divisor as it moves in its complete linear system on $X$. So for divisors which are hypersurface sections, you can look at the system of hypersurfaces in the ambient projective space.
It's also useful to note that in this situation, you can restrict-then-intersect or vice versa. So not only can you realize the deformation of $C$ by cutting with a different hyperplane to get a different conic that meets $C$ in two points (because $\mathcal O_C(1) \cong \mathcal O_{\mathbb {P^1}}(2)$), you can also instead look at two hyperplanes, which intersect in a line, and then intersect that line with $Q$ to get two points.
Some elements of reflection.
First of all, this answer of mine some years ago can be of some help. It deals with the determination of a ruled quadric knowing three of its 3 lines.
Preliminary material: let us recall how 2 families of skew lines generating a given hyperboloid with one sheet can be obtained.
Dealing with the example you give, with equation written under the form
$$y^2-z^2=1-x^2 \ \iff $$
$$\text{Surface} \ H \ : (y-z)(y+z)=(1-x)(1+x),\tag{1}$$
the two families of lines can be given the following equations:
$$\text{Lines} \ L_a : \ \begin{cases}y-z&=&a(1-x)\\y+z&=&\dfrac{1}{a}(1+x)\end{cases}\tag{2}$$
$$\text{Lines} \ L'_b : \ \begin{cases}y-z&=&b(1+x)\\y+z&=&\dfrac{1}{b}(1-x)\end{cases}\tag{3}$$
for any non-zero real number $a$ or $b$.
Remark: please note that, by multiplication of its 2 equations, (2) $\implies$ (1) ; implication of equations meaning inclusion of corresponding geometric entities ($\forall a, L_a \subset H$) as desired. For the same reason, $\forall b, L'_b \subset H$.
$H$ can be defined by selecting any 3 lines of a family, say $L_{a_1},L_{a_2},L_{a_3}$ or 2 in a family and the third in the other one.
But taking only two of these lines, for example $L_1$ and $L'_1$:
$$L_1 : \ \begin{cases}y-z&=&1-x\\y+z&=&1+x\end{cases} \ \text{and} \ L'_1 : \ \begin{cases}y-z&=&1+x\\y+z&=&1-x\end{cases}\tag{4}$$
whose intersection point is $(x,y,z)=(0,1,0)$ is not enough to define in a unique way a hyperboloid with one sheet (think for example, among other solutions, to the hyperboloid with one sheet symmetrical to $H$ with respect to the plane defined by $L_1 \cup L'_1$). Besides, there are other candidates: there exists Hyperbolic Paraboloids that can pass through these 2 lines.
Best Answer
The standard way is to use genus formula. For a curve $C\subset X$, $X$ a smooth projective surface, $2g(C)-2=C\cdot(C+K_X)$. So, in your case $-2=L\cdot(L+K_X)$. But, by adjunction, $K_X=O_X(d-4)$ and thus $L\cdot K_X=d-4$. So, we get, $L^2=2-d$.