I'll just tell you the equations for cutting out $\psi(C\times \mathbb P^2) $ from the Segre variety $S\subset \mathbb P^8$.
There is a similar result for $\psi(\mathbb P^2 \times D^*)$ and since $\psi(C \times D^*)=\psi(C\times \mathbb P^2)\cap \psi(\mathbb P^2 \times D^*)$ you will be done by adding your perfectly correct equation $z_{00} + z_{11} + z_{22}=0$ .
So let's find out $\psi(C\times \mathbb P^2)\subset S=\psi(\mathbb P^2\times \mathbb P^2)\subset \mathbb P^8 \:.$
The diabolical trick is to replace the equation $f(x,y,z)=0$ by the three equations $$f(x,y,z)\cdot a^2=0,f(x,y,z)\cdot b^2=0,f(x,y,z)\cdot c^2=0 $$ The equivalence of this system with $f(x,y,z)=0$ is due to the fact that we can't have simultaneously $a^2=b^2=c^2=0$.
The equations in the system can then be translated in equations in the variables $z_{ij}$.
Let's look at an example :
Suppose $f(x,y,z)=xy+yz+zx$. Then we say that $xy+yz+zx=0$ is equivalent to the system $$(xy+yz+zx)\cdot a^2=0,(xy+yz+zx)\cdot b^2=0,(xy+yz+zx)\cdot c^2=0 \quad (SYST) $$
Writing $(xy+yz+zx)\cdot a^2=axay+ayaz+azax$ etc. the above system $(SYST)$ becomes $$z_{00} z_{10} + z_{10}z_{20} +z_{20}z_{00}=0,z_{01} z_{11} + z_{11}z_{21} +z_{21}z_{01}=0, z_{02} z_{12} + z_{12}z_{22} +z_{22}z_{02}=0 $$ These three equations, joined of course to the equations $z_{ij}z_{kl} - z_{il}z_{kj}=0$ for the Segre embedding, define the subvariety $$\psi(C\times \mathbb P^2)\subset S=\psi(\mathbb P^2\times \mathbb P^2)\subset \mathbb P^8 \:.$$
(1) Let $e_1 = (1:0:0)$, $e_2 = (0:1:0)$, $e_3 = (0:0:1)$. Another way to phrase this claim is: given any $3$ points $\newcommand{\P}{\mathbb{P}} \newcommand{\PP}{\mathbb{P}} P_1, P_2, P_3 \in \mathbb{P}^2$ that are not collinear, there is a linear automorphism $\varphi: \P^2 \to \P^2$ that takes $P_i \mapsto e_i$ for $i = 1, 2, 3$. This can be proved using just a little linear algebra.
Write $P_1 = (a_0:a_1:a_2), P_2 = (b_0:b_1:b_2), P_3 = (c_0:c_1:c_2)$, and let $B$ be the matrix whose columns are $P_1, P_2, P_3$:
\begin{align*}
B =
\begin{pmatrix}
| & | & |\\
P_1 & P_2 & P_3\\
| & | & |
\end{pmatrix}
=
\begin{pmatrix}
a_0 & b_0 & c_0\\
a_1 & b_1 & c_1\\
a_2 & b_2 & c_2
\end{pmatrix} \, .
\end{align*}
Then $B$ takes $e_i$ to $P_i$. Assuming $P_1, P_2, P_3$ are not collinear, then $B$ is invertible. Letting $A = B^{-1}$, then $A$ takes $P_i$ to $e_i$ for all $i$. Thus the map
\begin{align*}
\varphi: \P^2 &\to \P^2\\
\begin{pmatrix}
x\\ y\\ z
\end{pmatrix}
&\mapsto
A
\begin{pmatrix}
x\\ y\\ z
\end{pmatrix} \, ,
\end{align*}
or written in coordinates,
$$
\varphi(x:y:z) = (A_{11} x + A_{12} y + A_{13} z : A_{21} x + A_{22} y + A_{23} z : A_{31} x + A_{32} y + A_{33} z) \, ,
$$
is an automorphism taking $P_i \mapsto e_i$ for $i = 1,2,3$. Or put differently, letting
\begin{align*}
X &= A_{11} x + A_{12} y + A_{13} z\\
Y &= A_{21} x + A_{22} y + A_{23} z\\
Z &= A_{31} x + A_{32} y + A_{33} z
\end{align*}
then the coordinates of $P_1, P_2, P_3$ with respect to the projective coordinates $X,Y,Z$ are $(1:0:0), (0:1:0), (0:0:1)$, respectively.
(2) We're trying to determine the coordinate ring of the affine conic $W \subseteq \mathbb{A}^2$. Considering $\mathbb{A}^2 \subseteq \P^2$ and taking the closure $\overline{W}$ of $W$ in $\P^2$, by the previous part, we have an isomorphism $\varphi: \overline{W} \overset{\sim}{\to} C$, where $C \subseteq \P^2$ is the conic $C: xy+yz+zx=0$. We can recover $W$ from $\overline{W}$ by removing the points at infinity, i.e., the points of intersection of $\overline{W}$ with the line $z=0$ at infinity. By Bézout's Theorem, $\overline{W}$ intersects $z=0$ in $2$ points, counted with multiplicity, so either $1$ or $2$ points. Thus $W = \overline{W} \setminus \{Q_1, Q_2\}$ or $W = \overline{W} \setminus \{Q_1\}$. Letting $R_i = \varphi(Q_i)$ for $i=1,2$, then restricting $\varphi$ yields an isomorphism $W \overset{\sim}{\to} C \setminus \{R_1, R_2\}$ or $W \overset{\sim}{\to} C \setminus \{R_1\}$.
Now over an algebraically closed field, every conic is isomorphic to $\P^1$. (Fix a point on the conic and consider the pencil of lines through this point, or see this post.) Denote this isomorphism by $\psi: C \overset{\sim}{\to} \P^1$. By post-composing with an automorphism of $\P^1$, we can assume that $R_1, R_2$ map to $\infty, 0$, or $R_1$ maps to $\infty$ under $\psi$. Composing the restrictions of $\varphi$ and $\psi$, we obtain either an isomorphism $\psi \circ \varphi: W \overset{\sim}{\to} \P^1 \setminus \{0, \infty\} \cong \mathbb{A}^1 \setminus \{0\}$ or $\psi \circ \varphi: W \overset{\sim}{\to} \P^1 \setminus \{\infty\} \cong \mathbb{A}^1$. These varieties have coordinate rings $k[x,1/x]$ and $k[x]$, respectively.
(3) Yes, this is true, although a degree $2$ hypersurface in $\P^n$ is usually called a quadric hypersurface, not a conic, when $n \geq 3$. See this post for a proof.
Best Answer
A degenerate conic is given by an equation that factors: that is, the conic $V(ax^2+bxy+cy^2+dxz+eyz+fz^2)$ is degenerate iff $$ax^2+bxy+cy^2+dxz+eyz+fz^2 = (px+qy+rz)(sx+ty+uz).$$ Now, just as you identified the six coefficients of the equation of the conic with points in $\Bbb P^5$, you can identify the coefficients on each of the linear equations with points in $\Bbb P^2$. Expanding the product $$(px+qy+rz)(sx+ty+uz) = (ps)x^2+(pt+qs)xy+(qt)y^2+(pu+rs)xz+(qu+rt)yz+(ru)z^2$$ we see that all the entries on the right hand side are sums of products of one coordinate from the first entry and one coordinate from the second entry. As these products are exactly the coordinates on the Segre embedding and we can take sums of coordinates through a linear projection, we have the result. (I'd recommend working through an example to see how this projection works for yourself - try projection from $[0:-1:1]$ to $V(x_2)$ in $\Bbb P^2$ and note that a point $[a:b:c]$ goes to $[a:b+c:0]$).