Consider the torus $S^1 \times S^1$ embedded in $\mathbb{R}^3$, defined implicitly as $$\mathbb{T} = \{(x,y,z) \in \mathbb{R}^3 \ \mid \ \left( \sqrt{x^2+y^2}-2 \right)^2 + z^2 = 1 \}. $$ Endow $\mathbb{T}$ with the subspace topology of $\mathbb{R}^3$. We know that $\mathbb{T}$ is a $2$-dimensional embedded submanifold of $\mathbb{R}^3$.
Now consider $p = (1,0,0) \in \mathbb{T}.$ I want to find a local parametrization of $\mathbb{T}$ around $p$, that is, a function $$f: U \to V, $$ where $V \subset \mathbb{T}$ is open in $\mathbb{T}$ and $p \in V$, $U$ is an open set of $\mathbb{R}^2$ and $f$ is a diffeomorphism (or at least a homeomorphism), and after that, find a correspoding chart and adapted chart to $\mathbb{T}$.
Of course, I thought about the standard parametrization of the torus, that is, the function $$f: (\theta, \phi) \mapsto ((2+\cos(\theta))\cos(\phi), (2+\cos(\theta))\sin(\phi), \sin(\theta)). $$
However, how can I find an actual open set of $\mathbb{R}^2$ and an open set of $\mathbb{T}$ such that the above properties hold? We can see that, with the above function, $f(\pi, 0) = (1,0,0) = p.$ Hence, I thought about a neighbourhood of $(\pi, 0)$, namely $\displaystyle U = \left( \frac{3\pi}{4}, \frac{5\pi}{4} \right) \times \left( -\frac{\pi}{4}, \frac{\pi}{4} \right) \subset \mathbb{R}^2.$ Obviously this is an open set of $\mathbb{R}^2$. After that, I though about proving that $V = f(U)$ will be open in $\mathbb{T}$ (I chose this $V$ so that $f$ is bijective), but I am not sure how.
In doing so, I thought about the chart of $\mathbb{T}$ around $p$ corresponding to $f$, by which I mean $(V, \varphi)$, with $\varphi: V \to U, \varphi = f^{-1}$. I will remark that it is trivial to see that $f$ is smooth on $U$. Furthermore, with this definition, we get that $$\varphi (x,y,z) = \left( \arcsin(z), \arctan \left( \frac{y}{x} \right) \right).$$
The function $\varphi$ is smooth if it is defined on a set that does not contain points $(x,y,z)$ with the property that $x = 0$. Particularly, these type of points are not in $V$ by the definition of $U$ and $f$, so $\varphi$ is smooth, hence, in particular, $f$ is a homeomorphism, so $V$ is an open set of $\mathbb{T}$. Is this good enough for the question?
Now, I want to find an adapted chart to $\mathbb{T}$ around $p$, by which I mean a smooth chart of $\mathbb{R}^3$, $(X, \chi)$, with $X \subset \mathbb{R}^3$ open, $p \in X$, $\chi$ diffeomorphism between $X$ and another open subset of $\mathbb{R}^3$ and $$\mathbb{T} \cap X = \{t \in X \ \mid \ \chi_3(t) = 0 \}, $$ where $\chi = (\chi_1, \chi_2, \chi_3)$ is the coordinate representation of $\chi$.
However, I don't know how to construct such an adapted chart to $\mathbb{T}$. I know we can somehow determine it by using the previous parametrization $f$ and chart $\varphi$, but I don't know how.
Edit:
I have thought a bit more about the last question. Observe that by letting $g: V \to \mathbb{R}, g(x,y,z) = 1 - \left( \sqrt{x^2+y^2}-2 \right)^2 - z^2$, we have that $g$ is smooth on $V$. Can we deduce that the function $\chi' = (\varphi, g)$ is such an adapted chart of $\mathbb{T}$?
Edit 2:
The function $\chi'$ would satisfy the property that $\mathbb{T}$ can be locally written as $g = 0$, but I cannot find a good domain and range to define the function such that it is an actual chart (bijective, smooth, with smoot inverse). More specifically, it's not hard to choose a domain such that $\chi'$ is smooth, but how do I choose it such that it is bijective, because $g$ is a function from $\mathbb{R}^3$ to $\mathbb{R}$, which is difficult to make injective.