I don't know how Lee explains this, but as soon as we pick local coordinates $x^i$ we have 3 related (but distinct) things:
Coordinate functions $x^i$.
Coordinate vector fields $\partial_i = \partial/\partial x^i$.
Coordinate covector felds $dx^i$.
These satisfy a bunch of relations, e.g. $\partial_j x^i = \delta_j^i$ (Kronecker delta) and $dx^i(\partial_j) = \delta_j^i$.
If we have a map $x = \phi(y)$ given in coordinates by $x^i = \phi^i(y^1, \cdots, y^m)$, there is a notion of pushforward of a tangent vector. If we have some tangent vector (in the $y$ coordinates which is given by
$$ v = \sum_i v^i \frac{\partial}{\partial y^i} $$
then we obtain a vector $\phi_\ast v$ in the $x$ coordinates by
$$ \phi_\ast v = \sum_{ij} v^i
\frac{\partial \phi^j}{\partial y^i} \frac{\partial}{\partial x^j}. $$
If you think of tangent vectors as being derivations of smooth functions, this is completely obvious: it's just the chain rule.
Ok, so if you understand the chain rule, then you understand pushforwards, and if you understand pushforwards you understand pullbacks, since they are just dual to pushforwards. Remember that a $p$-form is just a (skew) multilinear function. So to define a $p$-form in the $y$ coordinates you just have to say how to act on tangent vectors. But we already know how to push tangent vectors forward to the $x$ coordinates. So if $\omega$ is a $p$-form on in the $x$-coordinates, we obtain a $p$-form in the $y$-coordinates by the formula
$$ (\phi^\ast \omega)_y(v_1, \cdots, v_p) = \omega_{\phi(y)}(\phi_\ast v_1, \cdots, \phi_\ast v_p) $$
as you wrote above. To emphasize: this is all just the chain rule and linear algebra.
Now let's look at your particular example. We start with coordinates $x,y$ and want to pass to coordinates $r,t$, via
$$ x = r \cos t $$
$$ y = r \sin t $$
Call this map $(r,t) \mapsto (x,y)$ $\phi$.
So, how do we push vectors forward? By the chain rule, we have
\begin{align}
\phi_\ast(\frac{\partial}{\partial r})
&= \frac{\partial x}{\partial r} \frac{\partial}{\partial x}
+ \frac{\partial y}{\partial r} \frac{\partial}{\partial y} \\
&= \cos t \frac{\partial}{\partial x} + \sin t \frac{\partial}{\partial y}
\end{align}
Similarly,
\begin{align}
\phi_\ast(\frac{\partial}{\partial t})
&= \frac{\partial x}{\partial t} \frac{\partial}{\partial x}
+ \frac{\partial y}{\partial t} \frac{\partial}{\partial y} \\
&= -r\sin t \frac{\partial}{\partial x} + r \cos t \frac{\partial}{\partial y}
\end{align}
Now, to get the pullback of $dx \wedge dy$, we just need to evaluate it on these pushforwards (now edited to include more detail):
\begin{align}
(dx \wedge dy)(\phi_\ast\partial_r, \phi_\ast\partial_t) &=
(dx \wedge dy)(\cos t \partial_x + \sin t \partial_y, -r\sin t \partial_x + r\cos t \partial_y) \\
&= r\cos^2 t (dx\wedge dy)(\partial_x, \partial_y) -r\sin^2 t (dx\wedge dy)(\partial_y, \partial_x) \\
&= r\cos^2 t + r \sin^2 t \\
&= r
\end{align}
Here, in going from the first line to the second I used the bilinearity and skew-symmetry properties of forms (e.g. $(dx\wedge dx)(\partial_x, \partial_x) = 0)$) to expand it. To go from the second to the third I used the fact that $(dx\wedge dy)(\partial_x, \partial_y) = 1$, which is true because $dx$ and $dy$ are dual to $\partial_x$ and $\partial_y$.
That is,
$$ \phi^\ast (dx \wedge dy)(\partial_r, \partial_t) = r $$
So we have
$$ \phi^\ast (dx \wedge dy) = r dr \wedge dt. $$
Note that this is the same result you'd obtain if you just differentiated $x = r\cos t$ and $y = r\sin t$ and then wedged the results together. The reason this works is because of two facts:
Pullbacks commute with the exterior derivative $d$.
The exterior derivative of a coordinate function $x^i$ is exactly the coorinate covector field $dx^i$. (Hence the notation!)
For any $p=(x, y, z, t)$ in $\mathbb S^3$, the vector
$$ v =(y, -x, t, -z)= y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y} + t\frac{\partial}{\partial z}-z\frac{\partial}{\partial t}$$
is in $T_p\mathbb S^3$ since it's perpendicular to the normal vector $(x, y, z, t)$. Another way to see that: consider the curve
$$\gamma (t) = \sin t (y, -x, t, -z) + \cos t (x, y, z, t).$$
Since $ (y, -x, t, -z)$ is perpendicular to $ (x, y, z, t)$, $|\gamma (t)|^2 =1$. Thus $\gamma(t)$ is a curve in $\mathbb S^3$ and $\gamma(0) = p$. Then $\gamma'(0) =(y, -x, t, -z)$ is an element in $T_p\mathbb S^3$.
Then
$$\alpha (v) = -y^2 -x^2 - t^2 - z^2 = -1\neq 0$$
Thus $\alpha$ does not vanish at any point.
Best Answer
No $dt$ — you are doing the pullback by one fixed group element at a time, so $t$ is fixed.