Integration of a $2$-form

I've never seen a definition of the integral of a $2$-form along a "2D path" $C: I^2 \to \mathbb{R}^2$ (has anyone?), but it seems clear to me that the sensible definition should be

$$ \int_C \omega := \int_{I^2} C^*\omega$$

where $C^*\omega$ is the pull-back of $\omega$ by $C$. Recall that it is defined by $$ (C^*\omega)_{|p} (u, v) := \omega_{|C(p)}(dC_{|p} (u), dC_{|p} (v))~.$$

NB: More generally, we could define the same way integrals $\int_C \omega$ where $\omega$ is a $k$-form on a $k$-dimensional manifold $M$ and $c$ is a smooth map $U \subset \mathbb{R}^k \to M$.

Let's come back to your problem. Practically, you can compute $C^* \omega$ by letting $(x,y) = C(t_1,t_2)$ in the expression of $\omega$, you get:

$$ \begin{align*} x &= (t_1 + 1) \cos (2\pi t_2)\\ y &= (t_1 + 1) \sin (2\pi t_2) \\ \end{align*} $$

hence

$$ \begin{align*} x^2 + y^2 &= (t_1+1)^2\\ dx &= \cos (2\pi t_2)\,dt_1 - 2\pi (t_1 + 1)\sin(2\pi t_2) dt_2\\ dy &= \sin (2\pi t_2)\,dt_1 + 2\pi (t_1 + 1)\cos(2\pi t_2) dt_2\\ dx \wedge dy &= 2\pi (t_1+1)\, dt_1\wedge dt_2\\ \end{align*} $$ thus $$ C^* \omega = {2\pi\, dt_1\wedge dt_2\over t_1 +1} $$ You can now easily compute your integral: assuming $I = [0,1]$ (you don't say what I is) $$ \int_C \omega = \int_{I^2}{2\pi\, dt_1\wedge dt_2\over t_1 +1} = \int_{I^2}{2\pi\, dt_1 dt_2\over t_1 +1} $$ which gives us by Tonelli's theorem $$ \int_C \omega = 2\pi\left(\int_{0}^1{dt_1\over t_1 +1}\right)\left(\int_{0}^1 dt_2\right) = 2\pi \log (2) $$

NB: Another approach ("just for fun") would be to work in polar coordinates, I'll let you try to figure out how do that.