By trial and error I found that $dx,dy$ are generators of $H^1_{dR}$ of $T=S^1\times S^1$. Verifying that they generate the first cohomology group is not difficult.
My problem is: I found them by trying some $1$-forms and verifying their properties but I would really like to have geometric insight and be able to derive why these generate the first cohomology group.
My work/idea and the problems I ran into:
Parameterise $S^1 \times S^1$ with the angles $(x, y)\in [0,2\pi]^2$. Normally we choose local coordinates because global coordinates do not exist but in this case they do exist.
Problem: It is not clear to me that if a $1$-manifold $M$ is parameterized by a global coordinate $x$ and another $1$-manifold $M'$ is parameterised by a coordinate $y$ then the product $M \times M'$ is parameterized (again globally) by $(x,y)$.
But for the sake of this question let's assume that indeed the product is parameterized by the product of the coordinates.
My real problem is:
How do I obtain the differential $1$-forms $dx,dy$ from the angles $x,y$?
My idea was to take the total derivative of the angle functions. But if we denote the angle by $\theta$ instead of $x$ then the total derivative turns out to be $$ d\theta = -{y \over x^2 + y^2}dx + {x \over x^2 + y^2}dy$$
which leaves me confused.
The coordinates $x$ and $y$ are not global. If they were, the torus would be diffeomorphic to $\mathbb{R}^2$. If you are thinking of $S^1$ as $[0, 2\pi]$ where the endpoints are identified, then $x$ and $y$ are coordinates on $(0, 2\pi)\times(0, 2\pi)$ which is the complement of the red and purple circles in the image below.
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The question still remains, how do you obtain the one-forms $dx$ and $dy$? They are obtained in the same way: $dx$ comes from the first factor and $dy$ comes from the second factor, so we just need to see how to obtain $dx$ on $S^1$. Consider $S^1$ as $\mathbb{R}/\mathbb{Z}$, the quotient of $\mathbb{R}$ by the equivalence relation $p \sim p' \Leftrightarrow p - p' \in \mathbb{Z}$; we will denote the element of $\mathbb{R}/\mathbb{Z}$ corresponding to $p \in \mathbb{R}$ by $[p]$.
Now, $dx$ is a well-defined one-form on $\mathbb{R}$, but it also descends to a one-form on $\mathbb{R}/\mathbb{Z}$. We define the one-form $dx$ on $\mathbb{R}/\mathbb{Z}$ by $dx_{[p]} := dx_p$. This is well-defined because if $p \sim p'$, then $dx_p = dx_{p'}$ (in fact $dx_p = dx_q$ for any $p$ and $q$). Despite the fact that the one-form we obtain on the torus is denoted $dx$, it is not an exact form - I discuss this further in my answer to your new question.
If you would prefer to think of $S^1$ as $[0, 2\pi]$ where $0$ and $2\pi$ are identified (i.e. $S^1$ parameterised by angle), then the above discussion holds for $\mathbb{R}/2\pi\mathbb{Z}$.
The final formula you use is when $\theta$ is the second polar coordinate on $\mathbb{R}^2\setminus\{0\}$ and $x$ and $y$ are the usual Euclidean coordinates. Here $x$ is a coordinate of the first copy of $S^1$ and $y$ is a coordinate on the second copy.