We can use the first equation to express $y$ in terms of $x^2$, or indeed $x$ in terms of $y$. Expressing $x$ in terms of $y$ looks more complicated, so solving for $y$ in terms of $x$ sounds better. It involves a fraction of course, and fractions are broken numbers, nice to avoid if possible.
If we scan the second equation, we can see that we can express $x$ in terms of $y^2$ in a simple way.
The second equation is equivalent to the equation $x=2y^2$.
Thus the two equations together are equivalent to the simpler system of equations
$$3(2y^2)^2 -12y=0,\qquad x=2y^2.$$
The first of these equations is equivalent to $y(y^3-1)=0$. Since $y^3-1=(y-109y^2+y+1)$, and $y^2+y+1=0$ has no real solutions, the real solutions of $y(y^3-1)=0$ are $y=0$ and $y=1$. There are also a couple of complex non-real solutions that I am not sure you are expected to look for.
Now that the possible values of $y$ have been determined, we use $x=2y^2$ to find the accompanying values of $x$.
Remark: Note that during the solution process, we were always replacing systems of equations by equivalent systems. Thus no roots have been lost, and none of the roots we have found can be spurious.
In solving equations or systems, we often do not bother to work with equivalent systems. Often the logic goes thus. If $x$ (or $(x,y)$) is a solution of Equation (or System) $1$, then it is a solution of Equation $2$. And if that is so, it is a solution of simple Equation $3$. We solve Equation $3$, getting perhaps several solutions. But then we must substitute in the original equation or system to check, for each candidate, whether it really is a solution. This is because some manipulations, most importantly squaring, can introduce spurious roots.
You're correct that his presentation is a bit confusing. He's using $x$ both for the original coordinate (in the $xyz$ coordinate system) and to represent the new coordinate along $L$ in the plane $P$. So to clarify, let's use $x'$ as the indicated (signed) distance from the point $Q$ to the point (which I'll call $R$ for now) on $L$. Then Spivak's claim is that the original $xz$-coordinates of $R$ are linear expressions in $x'$. Let $S$ be the point on the $x$-axis directly below $R$.
If you'll allow me to introduce a little trigonometry for a moment, let $\theta$ be the angle between the two lines, and let $Q=(\beta,0)$. Then $\alpha=\cos\theta = QS/QR$, so $QS = \alpha(QR) = \alpha x'$. Since $S=(x,0)$, we have $x-\beta = \alpha x'$, so $x=\alpha x' + \beta$. Of course, we do not need the definition of the cosine function to observe that as $R$ moves along the line $L$, the ratio of the signed distances $QS/QR$ stays constant; this is just similar triangles.
Now proceed with his argument, substituting $x=\alpha x'+\beta$. The $x'y$-coordinates in the plane $P$ are, of course, not the usual $xy$-coordinates coming from $3$-space.
Last comment: Since the equation of $L$ is $z=Mx+B$, then we have $\beta = -B/M$ and $M=\tan\theta$, so $\alpha=\cos\theta = (\pm)\frac1{\sqrt{M^2+1}}$.
Best Answer
At the point $(x_0, y_0)$ where they intersect, $(x_0, y_0)$ satisfies both equations, that is: $$ \left\{\begin{array}{rcl}y_0 = 2x_0 + 4 \\ y_0 = 3x_0 + 5 \end{array}\right. $$ In particular, both of the right-hand sides are equal to $y_0$, so they are equal to one another: $$2x_0 + 4 = 3x_0 +5.$$ Now, rearranging gives $$x_0 = -1.$$ Again, the point $(x_0, y_0)$ is on both lines, so we can substitute it in either equation, say, the first, and get a value for $y_0$: $$y_0 = 2(-1) + 4 = 2.$$ So, the intersection point is $$(x_0, y_0) = (-1, 2),$$ and we can check this, if we like, by substituting in the other equation.
It may be an instructive exercise, by the way, to work this out for a general pair of lines in the plane specified this way, that is, for the lines $$ \left\{\begin{array}{rcl}y &=& m\phantom{'} x + b\phantom{'} \\ y &=& m' x + b' \end{array}\right. $$ Note that when $m = m'$ there are either no solutions or infinitely many solutions---what do these special cases correspond to geometrically?