EDIT: After having posted the question along with an attempt at a solution, I could spot the mistake in my solution (it's easier to read LaTeX than your own imperfect handwriting...), so now it is solved and I will post my solution as an answer, while I only leave the main question in the main post. Of course I'm more than happy to accept an alternative solution.
I'll be referring to Lee's Intro to Smooth Manifolds book whenever citing a result. I'm tackling Problem 16-2.
Consider the torus $\mathbb T^2=S^1\times S^1\subset\mathbb R^4$. Write an arbitrary element of $\mathbb R^4$ as $(x^1,x^2,x^3,x^4)$. Consider the $2$-form $\omega=x^2 x^3x^4 dx^1\wedge dx^3$. I need to compute $$ \int_{\mathbb T^2}\omega. $$ Let $p:(0,2/pi)\to S^1-\{(0,1)\},t\mapsto e^{it}$. Consider the parametrization $$ p\times p:(0,2\pi)\times(0,2\pi)\to \mathbb T^2,(t,s)\mapsto (e^{it},e^{is}). $$ I would like to show that this parametrization is orientation preserving, so that I can invoke Proposition 16.8. However, I have difficulty showing this.
By proposition 15.7 we can consider orientations forms $\omega_1,\omega_2$ on $S^1$, and then $\pi_1^*\omega_1\wedge\pi_2^*\omega_2$ will be an orientation form on $S^1\times S^1$. Since $S^1$ is a hypersurface in $\mathbb R^2$, we know by Proposition 15.21 that if we take an orientation form $\eta_1$ on $\mathbb R^2$, then $\iota^*(N\lrcorner\omega)$ will be an orientation form of $S^1$. Take $dx^1\wedge dx^2$ as orientation form on $\mathbb R^2$. Then $$ \iota^*(N\lrcorner dx^1\wedge dx^2)=x^1 dx^2-x^2 dx^1. $$ Similarly we find an orientation form for the other factor of the torus. This then yields $$ \pi_1^*(x^1 dx^2-x^2 dx^1)\wedge \pi_2^*(x^3 dx^4-x^4 dx^3)=(x^1 dx^2-x^2 dx^1)\wedge (x^3 dx^4-x^4 dx^3) $$ as orientation form on $S^1\times S^1$. Let's call this form $\eta$. Using coordinates $(t,s)$ on $(0,2\pi)^2$, if we pull back along $p\times p$ we get $$ (p\times p)^*\eta=(\cos t\, d\sin t-\sin t\, d\cos t)\wedge(\cos s\, d\sin s-\sin s\, d\cos s) $$ which is equal to $$ (\cos^2 t+\sin^2 t)(\cos^2 s+\sin^2 s)dt\wedge ds=dt\wedge ds. $$ Note that this is an orientation form on $(0,2\pi)\times (0,2\pi)$ and therefore $p\times p$ is orientation preserving by Exercise 15.13. Computing the integral now is straightforward.
Cleaner solution:
Here is a similar solution using that $p$ is a coordinate chart. Let's first describe an orientation form on $S^1$ in terms of the chart $p$. Note that $p$ has to be either orientation preserving or reversing, since $(0,2\pi)$ is connected. Consider the oriented frame $\partial_t$ on $(0,2\pi)$. We have that $$ T_tp \partial_t=-\sin t \partial_x + \cos t\partial_y, $$ which is positively oriented (by Prop 15.21 a frame $X$ for $S^1$ is oriented iff $(N,X)$ is oriented), and hence $p$ is orientation preserving. Let $\theta=p^{-1}$. Then $d\theta=(p^{-1})^*dt$, which is by construction an orientation form. Denoting the same form on the other factor of the torus by $d\phi$, we find that $d\theta\wedge d\phi$ is an orientation form on $S^1\times S^1$, and by construction we have that $$ (p\times p)^*d\theta\wedge d\psi=dt\wedge ds $$ is an orientation form, and hence $p\times p$ is orientation preserving. (Note that $p\times p$ would have been orientation preserving even if $p$ was orientation reserving, but it's nice to know that $p$ is orientation preserving - as it should be!)