Hello, back from a long drive and dinner.
There are five glaring mistakes on page 2, which is exactly what we expect when the level of discussion is "the article used calculus"
The Five mistakes I would like to unravel are:
1) Incorrect manipulation of expressions, which ignore the postulates. Thus he stops talking about the physics of light.
2) Ignoring the rest of the Lorentz transform
3) The absurd assertion that radially propagating light also propagates along a vertical line with fixed y.
4) The misuse of multivariable calculus that asserts that just because "$$\frac{\partial x'}{\partial x} > 0$$, if x increases, then x' increased."
5) This misuse of multivariable calculus that asserts that $$\frac{\partial x'}{\partial x} > 0$$ and $$\frac{\partial x}{\partial t} > 0$$ and $$x' < 0$$ are enough to conclude that in the primed frame the light is moving toward the coordinate origin.
Correct manipulation of expressions
Given $$z = 0 \, , \; 0 < x \, , \; 0 < y \, , \; 0 < v < c $$, $$c^2 t^2 = x^2 + y^2$$ describes a two-dimensional surface. So solutions are necessarily parameterized by two independent variables.
$$t(x,y) = \frac{1}{c} \sqrt{x^2 + y^2}$$ is one possible parameterization.
$$x(\theta,t) = ct \, \cos \theta \, , \; y(\theta,t) = ct \, \sin \theta$$ is another.
$$x(y,t) = \sqrt{c^2 t^2 - y^2}$$ is another.
So when we parameterize x', t' and y' in terms of x and t, we have the related choices:
$$x'(x,y) = \left( x - v t(x,y) \right) \gamma = \frac{c x - v \sqrt{x^2 + y^2}}{c \sqrt{1 - \frac{v^2}{c^2}}} = \frac{c x - v \sqrt{x^2 + y^2}}{\sqrt{c^2 - v^2}} \\ y'(x,y) = y \\ t'(x,y) = \left( t(x,y) - \frac{v x}{c^2} \right) \gamma = \frac{\frac{1}{c} \sqrt{x^2 + y^2} - \frac{v x}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{c \sqrt{x^2+y^2}-v x}{c \sqrt{c^2-v^2}} $$
Partial derivatives are only partial derivatives
Now we can take partial derivatives of these expressions, remembering to do the work Mr Banks ignored:
$$ \begin{array}{rclrclrclrcl} \frac{\partial x'(x,y)}{\partial x} & = & \frac{c - \frac{v x}{\sqrt{x^2 + y^2}}}{\sqrt{c^2 - v^2}} & \frac{\partial y'(x,y)}{\partial x} & = & 0 & \frac{\partial t'(x,y)}{\partial x} & = & \frac{\frac{cx}{\sqrt{x^2 + y^2}} - v}{c \sqrt{c^2 - v^2}} & \frac{\partial t(x,y)}{\partial x} & = & \frac{x}{c\sqrt{x^2 + y^2}} \\ \frac{\partial x'(x,y)}{\partial y} & = & \frac{- v y}{\sqrt{x^2 + y^2}\sqrt{c^2 - v^2}} & \frac{\partial y'(x,y)}{\partial y} & = & 1 & \frac{\partial t'(x,y)}{\partial y} & = & \frac{y}{\sqrt{x^2 + y^2}\sqrt{c^2 - v^2}} & \frac{\partial t(x,y)}{\partial y} & = & \frac{y}{c\sqrt{x^2 + y^2}} \end{array}$$
So here we can see that even though $$\frac{\partial x'}{\partial x} > 0$$, $$\frac{\partial x'}{\partial y} < 0$$ and so the sign of $$\partial x'$$ is indeterminate unless we add a constraint on $$\partial x \, , \; \partial y$$. Mr. Banks sees to hold y constant, but then he is no longer describing the propagation of rays of light but where the expanding sphere of light meets the line $$y = y_g$$ which is akin to a searchlight beam hitting a wall and not akin to a particle. Like a spot on the wall lit up by a laser pointer, it is not constrained to move slower than light.
$$\frac{\partial x}{\partial t} = c \sqrt{1 + \frac{y^2_g}{x^2}} > c$$
Directions need more than one coordinate
Even with all the problems, does the imaginary choice move away from the new coordinate origin?
To answer this, we must take the dot product of the velocity with the position in the new coordinate system and check the sign.
$$x'(x,y) \frac{\partial x'}{\partial t'} + y'(x,y) \frac{\partial y'}{\partial t'} = x'(x,y) \frac{\partial x'}{\partial x} \frac{\partial x}{\partial t'} + y'(x,y) \frac{\partial y'}{\partial x} \frac{\partial x}{\partial t'} = c \frac{c \sqrt{x^2+y^2}-v x}{\sqrt{c^2-v^2}} $$ which is always greater than zero, indicating that even in the bad physics, Mr. Banks has also erred when "the article used calculus."
A better way
Because of the conservation of angular momentum, we should use a better parameterization to see the fate of actual rays of light. $$x(\theta,t) = ct \, \cos \theta \, , \; y(\theta,t) = ct \, \sin \theta$$ is natural.
So when we parameterize x', t' and y' in terms of theta and t, we have the related choices:
$$x'(\theta,t) = \left( x(\theta,t) - v t \right) \gamma = \frac{c \, \cos \theta - v}{\sqrt{1 - \frac{v^2}{c^2}}} t \\ y'(\theta,t) = y(\theta,t) = ct \, \sin \theta \\ t'(\theta,t) = \left( t - \frac{v x(\theta,t)}{c^2} \right) \gamma = \frac{1 - \frac{v}{c} \, \cos \theta }{\sqrt{1 - \frac{v^2}{c^2}}} t $$
$$ \begin{array}{rclrclrclrclrcl} \frac{\partial x'(\theta,t)}{\partial \theta} & = & \frac{-c \, \sin \theta }{\sqrt{1 - \frac{v^2}{c^2}}} t & \frac{\partial y'(\theta,t)}{\partial \theta} & = & ct \, \cos \theta & \frac{\partial t'(\theta,t)}{\partial \theta} & = & \frac{\frac{v}{c} \, \sin \theta }{\sqrt{1 - \frac{v^2}{c^2}}} t & \frac{\partial x(\theta,t)}{\partial \theta} & = & - c t \, \sin \theta & \frac{\partial y(\theta,t)}{\partial \theta} & = & c t \, \cos \theta \\ \frac{\partial x'(\theta,t)}{\partial t} & = & \frac{c \, \cos \theta - v}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{\partial y'(\theta,t)}{\partial t} & = & c \sin \theta & \frac{\partial t'(\theta,t)}{\partial t} & = & \frac{1 - \frac{v}{c} \, \cos \theta }{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{\partial x(\theta,t)}{\partial t} & = & c \, \cos \theta & \frac{\partial y(\theta,t)}{\partial t} & = & c \, \sin \theta \end{array}$$
So how fast is light moving in the unprimed frame (holding theta constant)?
$$\sqrt{\left( \frac{\partial x}{\partial t} \right)^2 + \left( \frac{\partial y}{\partial t} \right)^2} = c$$
And in the unprimed frame?
$$\sqrt{\left( \frac{\partial x'}{\partial t'} \right)^2 + \left( \frac{\partial y'}{\partial t'} \right)^2} = \sqrt{\left( \frac{\partial x'}{\partial t} \frac{\partial t}{\partial t'} \right)^2 + \left( \frac{\partial y'}{\partial t} \frac{\partial t}{\partial t'} \right)^2} = \sqrt{\left( \frac{c \, \cos \theta - v}{1 - \frac{v}{c} \, \cos \theta } \right)^2 + \left( c \sin \theta \frac{\sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{v}{c} \, \cos \theta } \right)^2} = c$$
So is the light moving towards the origin in the primed frame?
$$\left( \frac{c \, \cos \theta - v}{\sqrt{1 - \frac{v^2}{c^2}}} t \right) \left( \frac{c \, \cos \theta - v}{1 - \frac{v}{c} \, \cos \theta } \right) + \left( ct \, \sin \theta \right) \left( c \sin \theta \frac{\sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{v}{c} \, \cos \theta } \right) = \frac{c^2 t (v-c \cos \theta)^2}{\sqrt{c^2-v^2} (c-v \cos \theta)} + \frac{c^2 t \sqrt{c^2 - v^2} \sin^2 \theta}{c - v \cos \theta} = \frac{c^2 t (c-v \cos \theta)}{\sqrt{c^2-v^2}} > 0$$
Of course not. Only someone without the ability to understand special relativity would think so. Only someone who doesn't understand calculus would calculate so.
