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 $$\psi(C\times \mathbb P^2)\subset S=\psi(\mathbb P^2\times \mathbb P^2)\subset \mathbb P^8 \:.$$
Any line in $\mathbb{P}^{3}$ is an intersection of two planes. In order to understand what lines are inside the quadric $xw=yz$, it is enough to understand when the hyperplane sections of $xw=yz$ in $\mathbb{P}^3$ contain a line.
Let $ax+by+cz+dw=0$ be a plane in $\mathbb{P}^3$. By Bezout's theorem, the intersection of $ax+by+cz+dw=0$ with $xw=yz$ is a curve of degree 2. If this curve is irreducible, then there is no hope of finding a line there! If this curve is a union of two lines, then we have successfully found a pair of lines.
So we need to understand what conditions need to be imposed on the coefficients $[a, b, c, d]\in\mathbb{P}^3$ such that the plane $ax+by+cz+dw=0$ intersects $xw=yz$ in a pair of lines.
Claim. $ax+by+cz+dw=0$ intersects $xw=yz$ in a pair of lines if and only if $ad=bc$.
Proof. $(\Leftarrow)$ Multiply $ax+by+cz+dw=0$ by $w$ and use $xw=yz$ to get $ayz+byw+czw+dw^2=0$. This is a plane curve in $\mathbb{P}^{2}$ (with coordinates $y, z, w$), unsurprisingly. Now, multiply both sides by $b$ to get
$$
abyz+b^2yw+bczw+bdw^2=0
$$
Use the hypothesis $ad=bc$ to get
$$
abyz+b^2yw + adzw + bdw^2 = 0
$$
which conveniently factors as
$$
(az+bw)(by+dw)=0
$$
so we get a pair of lines.
$(\Rightarrow)$ I will leave this as an exercise. Try to factor the quadric equation, and show that such a factorization forces $ad=bc$. $\square$
So now we can answer the question "What are all the lines on the quadric surface $xw=yz$?" Well, the proof of the claim shows that the lines on $xw=yz$ are of the form $az+bw=0$ and $by+dw=0$ (viewed in $\mathbb{P}^2$ with coordinates $y,z,w$) such that $ad=bc$. Here $a, b, c, d$ come from the hyperplane $ax+by+cz+dw=0$. Now once you can fix any $[a, b]\in\mathbb{P}^{1}$, you get the line $az+bw=0$. And if you fix $[b, d]\in\mathbb{P}^{1}$, you get the line $by+dw=0$. I think these two families of lines are the desired rulings.
I am very interested in seeing a more concise and conceptual answer!
Best Answer
Observe, in the Segre embedding $\mathbb{P}\times\mathbb{P}\rightarrow\mathbb{P}^3$, we have $$ (x_0:x_1)\times(y_0:y_1)\mapsto(x_0y_0,x_0y_1,x_1y_0,x_1y_1). $$
If we choose the coordinates on $\mathbb{P}^3$ to be $(x,z,w,y)$, we see that $xy=zw$ on this surface.