It turns out that my question is all about linear algebra. As I mentioned in the chat, to find such a quadric surface, one needs to find a line $L'$ in $\mathbb P^4$, which lies in the normal direction of $L$ inside the cubic threefold $X$. Such a line is the image of a nonzero section $s$ in the normal bundle $N_{L|X}\cong \mathcal{O}\oplus \mathcal{O}$ and the quadric surface is determined by the family of disjoint lines $\cup_{t\in \mathbb C}ts(L)$.

In other words, by linearizing the local equation of the Fano variety of lines $F$ of $X$ in the Grassmannian $Gr(2,5)$ (i.e., throw out higher-order terms in 6.14 in Clemens-Griffiths), we can determine that the equation of $L'_t=ts(L)$ is

$$ \begin{cases} tax_1+x_2=0,\\ tbx_0-x_4=0,\\ tax_0-tbx_1-x_3=0. \end{cases}\tag{1}\label{1} $$

with $(a,b)\in \mathbb P^1$. By standard linear algebra, we cancel out $t$ and find that, if $(a,b)=(1,0)$ or $(0,1)$, the solution contains the quadric surface $$Q_1:x_4=0,~ x_2x_0+x_3x_1=0, \text{or}$$ $$Q_2:x_2=0,~x_4x_1+x_3x_0=0,$$ that mentioned in the question respectively. When both $a$ and $b$ are nonzero, the solution of $(\ref{1})$ contains a quadric surface $$a^2x_4+b^2x_2-abx_3=0,~ax_2x_0+ax_3x_1+bx_0x_1=0.\tag{2}\label{2}$$

Let $b\to 0$, one recovers $Q_1$. To recover $Q_2$, one cancel out $x_2$ by the linear equation in $(\ref{2})$ and let $a\to 0$.

It turns out that my question is all about linear algebra. As I mentioned in the chat, to find such a quadric surface, one needs to find a line $L'$ in $\mathbb P^4$, which lies in the normal direction of $L$ inside the cubic threefold $X$. Such a line is the image of a nonzero section $s$ in the normal bundle $N_{L|X}\cong \mathcal{O}\oplus \mathcal{O}$ and the quadric surface is determined by the family of disjoint lines $\cup_{t\in \mathbb C}ts(L)$.

In other words, by linearizing the local equation of the Fano variety of lines $F$ of $X$ in the Grassmannian $Gr(2,5)$ (i.e., throw out higher-order terms in 6.14 in Clemens-Griffiths), we can determine that the equation of $L'_t=ts(L)$ is

$$ \begin{cases} tax_1+x_2=0,\\ tbx_0-x_4=0,\\ tax_0-tbx_1-x_3=0. \end{cases}\tag{1}\label{1} $$

with $(a,b)\in \mathbb P^1$. By standard linear algebra, we cancel out $t$ and find that the $\mathbb P^{1}$-family of quadric surfaces are given by

$$ \begin{cases} \text{I}.~a^2x_4+b^2x_2-abx_3=0,\\ \text{II}.~ax_2x_0+ax_3x_1-bx_2x_1=0,\\ \text{III}. ~ax_4x_0-bx_4x_1-bx_3x_0=0. \end{cases}\tag{2}\label{2} $$

in $\mathbb P^1_{[a,b]}\times \mathbb P^4_{[x_0,...,x_4]}$, with one linear equation and two quadric equations. Note that the family is not a complete intersection: When $a\neq 0$, the equation $\text{III}$ is redundant while when $b\neq 0$, the equation $\text{II}$ is redundant. Moreover, the points $(a,b)=(1,0)$ and $(0,1)$ correspond to the two "obvious" quadric surfaces that mentioned in the question.

By the way, I finally fixed the typo in Clemens-Griffiths p.309, about the defining equation of the curve $B(\alpha_0,\alpha_1)$: the fourth equation should be u_5=\cdots=u_n=z_4=\cdots=z_n=0. (The original paper has a z_2 instead of z_4, which is wrong.) It turns out that the 1-parameter family of lines in Clemens-Griffiths is exactly the family $(\ref{1})$ we defined and $(\ref{2})$ are the explicit equations for the $\mathbb P^1$-family of quadric surfaces tangent to $X$ along $L$ mentioned in Lemma 6.18.

It turns out that my question is all about linear algebra. As I mentioned in the chat, to find such a quadric surface, one needs to find a line $L'$ in $\mathbb P^4$, which lies in the normal direction of $L$ inside the cubic threefold $X$. Such a line is the image of a nonzero section $s$ in the normal bundle $N_{L|X}\cong \mathcal{O}\oplus \mathcal{O}$ and the quadric surface is determined by the family of disjoint lines $\cup_{t\in \mathbb C}ts(L)$.

In other words, by linearizing the local equation of the Fano variety of lines $F$ of $X$ in the Grassmannian $Gr(2,5)$ (i.e., throw out higher-order terms in 6.14 in Clemens-Griffiths), we can determine that the equation of $L'_t=ts(L)$ is

$$ \begin{cases} tax_1+x_2=0,\\ tbx_0-x_4=0,\\ tax_0-tbx_1-x_3=0. \end{cases}\tag{1}\label{1} $$

with $(a,b)\in \mathbb P^1$. By standard linear algebra, we cancel out $t$ and find that, if $(a,b)=(1,0)$ or $(0,1)$, the solution contains the two quadric surfaces $$Q_1:x_4=0,~ x_2x_0+x_3x_1=0,$$ $$Q_2:x_2=0,~x_4x_1+x_3x_0=0,$$ that mentioned in the question. When both $a$ and $b$ nonzero, the solution of $(\ref{1})$ contains a quadric surface $$a^2x_4+b^2x_2-abx_3=0,~ax_2x_0+ax_3x_1+bx_0x_1=0.\tag{2}\label{2}$$

Let $b\to 0$, one recovers $Q_1$. To recover $Q_2$, one cancel out $x_2$ by the linear equation in $(\ref{2})$ and let $a\to 0$.

It turns out that my question is all about linear algebra. As I mentioned in the chat, to find such a quadric surface, one needs to find a line $L'$ in $\mathbb P^4$, which lies in the normal direction of $L$ inside the cubic threefold $X$. Such a line is the image of a nonzero section $s$ in the normal bundle $N_{L|X}\cong \mathcal{O}\oplus \mathcal{O}$ and the quadric surface is determined by the family of disjoint lines $\cup_{t\in \mathbb C}ts(L)$.

In other words, by linearizing the local equation of the Fano variety of lines $F$ of $X$ in the Grassmannian $Gr(2,5)$ (i.e., throw out higher-order terms in 6.14 in Clemens-Griffiths), we can determine that the equation of $L'_t=ts(L)$ is

$$ \begin{cases} tax_1+x_2=0,\\ tbx_0-x_4=0,\\ tax_0-tbx_1-x_3=0. \end{cases}\tag{1}\label{1} $$

with $(a,b)\in \mathbb P^1$. By standard linear algebra, we cancel out $t$ and find that, if $(a,b)=(1,0)$ or $(0,1)$, the solution contains the quadric surface $$Q_1:x_4=0,~ x_2x_0+x_3x_1=0, \text{or}$$ $$Q_2:x_2=0,~x_4x_1+x_3x_0=0,$$ that mentioned in the question respectively. When both $a$ and $b$ are nonzero, the solution of $(\ref{1})$ contains a quadric surface $$a^2x_4+b^2x_2-abx_3=0,~ax_2x_0+ax_3x_1+bx_0x_1=0.\tag{2}\label{2}$$

Let $b\to 0$, one recovers $Q_1$. To recover $Q_2$, one cancel out $x_2$ by the linear equation in $(\ref{2})$ and let $a\to 0$.