pullback of rational normal curve under Segre map

Question

Let $\nu:P^1 \rightarrow P^2$ be the veronese map of degree $2$, i.e. $[Y_0 : Y_1] \mapsto [Y_0^2 : Y_0 Y_1 : Y_1^2]$ and let $\sigma: P^1 \times P^2 \rightarrow P^5$ be the Segre map. Consider the image $\mathcal{X}$ of $\sigma \circ \nu$, i.e. the image of \begin{align} [X_0:X_1] \times[Y_0:Y_1]\stackrel{\nu}{\mapsto} [X_0:X_1]\times[Y_0^2 : Y_0 Y_1 : Y_1^2]\stackrel{\sigma}{\mapsto}[X_0 Y_0^2: X_0 Y_0 Y_1 : X_0 Y_1^2 : X_1 Y_0^2: X_1 Y_0 Y_1 : X_1 Y_1^2]. \end{align}

Then $\mathcal{X}$ is a projective variety of $P^5$ given by $7$ quadratic equations ($3$ for the Segre variety and $4$ for the image of the veronese variety).

Next, let $C_4$ be the rational normal curve, which arises as the image of $P^1 \rightarrow P^4$ given by \begin{align} [W_0 : W_1] \mapsto [W_0^4: W_0^3 W_1: W_0^2 W_1^2 : W_0 W_1^3 : W_1^4]. \end{align}

$C_4$ is given by $4$ quadratic equations.

Exercise 2.28 in Harris (AG-first course) says that $C_4$ can be realized as a hyperplane section of $\mathcal{X}$. I have been trying to see why this is true, by working with the pullback of $C_4$ under $\sigma$. More precisely, to prove the statement in Harris i would have to show that this pullback is given by the equation $Y_1^2 - Y_0Y_2=0$, which is the veronese variety, together with a bilinear form in $X_1,X_0$ (homogeneous coordinates of $P^1$) and $Y_0,Y_1,Y_2$ (homogeneous coordinates of $P^2$), which would correspond to a hyperplane in $P^5$.

When i pull back the equations of $C_4$ i get:

$(1): \, X_0^2(Y_1^2-Y_0 Y_2) = 0 \\ (2): \, X_0(X_0 Y_2^2 - X_1 Y_0 Y_1)=0 \\ (3): \, X_1(X_1 Y_0^2 - X_0 Y_1 Y_2)=0.$

Issue 1: The variety described by these equations does not entirely live in the veronese variety (since equation (1) is multiplied by $X_0^2$).

Issue 2: I am finding hard to believe that equations (1)-(3) can give the same variety of $P^1 \times P^2$ as the one given by $Y_1^2-Y_0 Y_2=0$ together with a bilinear form.

Question: How can these two issues be addressed?

PS: I will be happy to provide more details upon request; did not do so in the first place to avoid overloading the question. Also, general hints are good enough for me, no need to go into details.

******EDIT************************

For convenience i attach the description from Harris:

Answer 1

I remember that this exercise took me some hours of thinking as well ! I found a purely geometric solution that worked well if I am not mistaken : as a hint you can consider that $X_{2,2}$ is defined as the union of all the lines joining two isomorphic rational curves of degree 2 in two linear complementary disjoint subspaces of projective dimension 2 of $\mathbb{P^5}$.

For instance, if you want a solution a bit more analytic, you see easily that you can define $X_{2,2}$ as the set of points of $\mathbb{P}^5$ such that the matrix $$\begin{bmatrix} X_0 & X_1 & X_3 & X_4 \\ X_1 & X_2 & X_4 & X_5 \end{bmatrix}$$ has rank one (can you see why considering the geometric description I gave you ?). Then if you look into the hyperplane $[X_2=X_3]$ what do you see ? You can also intersect with any hyperplane of $\mathbb{P}^5$ not meeting any line of the ruling of the scroll $X_{2,2}$, you will see that you will always get a rational normal curve !