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The question reads:

Consider the torus $T^2 = S^1 \times S^1$, parametrized in $\mathbb{R}^3$ by$$\Gamma(\theta, \varphi) = (\cos(\theta)(R+r\cos(\varphi)), \sin(\theta)(R+r\cos(\varphi)), r\sin(\varphi))$$ where $0

Now I believe I understand the first part, just find the derivative of the map and then parametrize it into $x,y,z$, but I am totally confused about the second part. Am I supposed to compute the map on the dual space of the tangent vectors at $p$? If anyone could help it would be greatly appreciated. Thank you.

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Right, I think by the induced map you do it on the dual space, like this:

$I^*:T^*(T^2)\to T^*(\mathbb{R}^3), \alpha\mapsto\omega$

where $\Gamma^*\omega=\alpha$. That is $\alpha_p(\xi)=\omega_{\Gamma(p)}(\Gamma_*\xi)$ for $\xi\in T_pT^2$.

Now for $\wedge^2 T_p^*(T^2)\to \wedge^2 T_{\Gamma(p)}^*\mathbb{R}^3$ you use $I^*$ on the generators: $\alpha\wedge\beta\mapsto I^*\alpha\wedge I^*\beta$.

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    Would the basis for $\Lambda^2T^*_{\rho}$ be {$dy\wedge dz, dz\wedge dy, dx\wedge dy$}? – 2017-02-09
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    Right, that would be the basis for $\wedge^2 T^*\mathbb{R}^3$ and a basis for $\wedge^2 T^*(T^2)$ would be $d\theta\wedge d\varphi$. – 2017-02-09
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    Also I notice for the $x$-coordinate of $\Gamma(\theta,\varphi)$, do you mean $\cos\theta(R+r\cos\varphi)$? – 2017-02-09
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    Yes, most definitely. – 2017-02-09