projection of a quadric surface

The basic idea is that the "inverse map" is given by sending $\ell \in \mathbb{P}^2$ (identify the points in $\mathbb{P}^2$ with lines through $x$) to the point $y$ where $X \cap \ell = \{x, y\}$. (We are using that $X$ has degree 2, which means that it intersects a general line in 2 points.) It's pretty clear that this map is inverse to the one you described, wherever things are well-defined. It also should be clear that the sets of points in $X$ and $\mathbb{P}^2$ where the maps are not well-defined are proper closed subsets. Lastly, you need to check that these are actually morphisms where they are defined, and I'm afraid this step, by definition, requires a certain amount of coordinate computation to verify.

A side remark to help you understand the geometry of $X$ is that you can view it as the isomorphic image of $\mathbb{P}^1 \times \mathbb{P}^1$ under the Segre map $((a:b),(c:d))\mapsto (ac:bd:ad:bc)$ to $\mathbb{P}^3$. I chose a weird ordering for the products of the variables so that the equation $xy = zw$ would be satisfied, assuming you order your coordinates on $\mathbb{P}^3$ "alphabetically" as $(x:y:z:w)$. In particular, this shows that through every point of $X$ are exactly 2 lines in $\mathbb{P}^3$ contained in $X$. Indeed, the two copies of $\mathbb{P}^1$ give two separate rulings on $X$. Shafarevich's book has a nice discussion of this surface; you might also look at Igor Dolgachev's notes on Classical Algebraic Geometry.