Deriving generators for $H^1(T)$: what are $dx$ and $dy$?

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.