$\newcommand{\Blank}{\underline{\qquad}}$(Good-natured note: This isn't the first pullback you've computed. You've been computing pullbacks since you learned the chain rule and method of substitution.)
It's easiest to start by turning the clock back to 1850 or so. You have
\begin{align*}
x &= u - v, \\
y &= v^{2} - w,
\end{align*}
so the chain rule gives
\begin{align*}
dx &= \Blank\, du + \Blank\, dv + \Blank\, dw, \\
dy &= \Blank\, du + \Blank\, dv + \Blank\, dw.
\end{align*}
Now, to express this in modern terms, replace the $1$-forms on the left by pullbacks: $dx \to f^{*}dx$, etc.
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
I'll answer one of your questions in the comments, namely why $i^*(\alpha)$ is a non-vanishing 1-form on $S^3$.
Choose a point $p = (a, b, c, d) \in S^3$, so $i^*\alpha$ restricts to a linear functional on the tangent space $T_{p}(S^3)$. If $v \in T_p(S^3)$ is any tangent vector, then $v$ and $p$ (viewed as a vector in $\mathbb{R}^4$) must be perpendicular, i.e. we must have $$av_x + bv_y + cv_z + d v_t = 0$$ In particular, this implies that the vector \begin{align} v_1 := \langle -b, a, -d, c \rangle \end{align} is tangent to $S^3$ at $p$.
Now, observe that \begin{align} i^*(\alpha)|_p(v_1) &= a (a) - b (-b) + c(c) - d (-d) = a^2 + b^2 + c^2 + d^2 = 1 \neq 0. \end{align} Therefore $i^*(\alpha)|_p$ is not the zero functional. This argument works for any point $p$.
Remark: this argument is essentially saying that the vector field given by $v_1$ is non-vanishing as a vector field on $S^3$.