The answer is positive. The maximal atlas containing an arbitrary atlas is obtained by adding all the charts that are compatible with the atlas. As pointed out by Gnampfissimo in the comment, the crux of the problem lies in showing that if two charts, say $(V_1,\psi_1 )$ and $(V_2,\psi_2 )$, are compatible with an atlas $\mathfrak{U}$, then $(V_1,\psi_1 )$ and $(V_2,\psi_2 )$ are compatible with each other. So let us try to prove this.
Let $p\in V_1\cap V_2$. We would like to show that $\psi_2\circ\psi_1^{-1}:\psi_1(V_1\cap V_2)\rightarrow\psi_2(V_1\cap V_2)$ is smooth at $\psi_1(p)$. To prove this, choose a chart $(U,\phi )$ in $\mathfrak{U}$ that contains $p$. Then both $\phi\circ\psi_1^{-1}:\psi_1(U\cap V_1)\rightarrow\phi(U\cap V_1)$ and $\psi_2\circ\phi^{-1}:\phi(U\cap V_2)\rightarrow \psi_2(U\cap V_2)$ are smooth. So there are
- neighborhoods $O_1$, $O_2$ of $\psi_1(p)$ and $\phi(p)$, respectively, in $\mathbb{R}^n$,
- a smooth map $f:O_1\rightarrow\mathbb{R}^n$ that agrees with $\phi\circ\psi_1^{-1}$ on $O_1\cap\psi_1(U\cap V_1)$, and
- a smooth map $g:O_2\rightarrow\mathbb{R}^n$ that agrees with $\psi_2\circ\phi^{-1}$ on $O_2\cap\phi(U\cap V_2)$.
By taking the intersection of $O_1$ with $f^{-1}(O_2)$ if necessary, we may assume here that $O_1$ satisfies $f(O_1)\subset O_2$. Then $g\circ f:O_1\rightarrow \mathbb{R}^n$ is a smooth map from a neighborhood of $\psi_1(p)$ to $\mathbb{R}^{n}$. So if we can show that $g\circ f$ agrees with $\psi_2\circ\psi_1^{-1}$ on $O_1\cap\psi_1(V_1\cap V_2)$, then we can say that $\psi_2\circ\psi_1^{-1}:\psi_1(V_1\cap V_2)\rightarrow\psi_2(V_1\cap V_2)$ is smooth at $\psi_1(p)$, and we are done.
For sure, $g\circ f$ agrees with $\psi_2\circ\psi_1^{-1}$ on $O_1\cap\psi_1(U\cap V_1\cap V_2)$. But what about on the set $\{O_1\cap \psi_1(V_1\cap V_2)\}- \{O_1\cap \psi_1(U\cap V_1\cap V_2)\}$? We have no information whatsoever about the behavior of $g\circ f$ on this set. What to do?
Here's a way to proceed: Since $\psi_1(U\cap V_1 \cap V_2)$ is open in $\psi_1(V_1\cap V_2)$, there is an open set $A$ in $\mathbb{R}^n$ such that $A\cap \psi_1(V_1\cap V_2)=\psi_1(U\cap V_1 \cap V_2)$.
Then we have $$(O_1\cap A)\cap \psi_1(U\cap V_1\cap V_2)=(O_1\cap A)\cap \psi_1 (V_1\cap V_2),$$
so the replacement of $O_1$ by $O_1\cap A$ fixes everything.
For Question 1, you are right. For instance, you can just take the set of all charts on $(M,T)$ and they will be an atlas.
For Questions 2 and 3, as you have defined a topological manifold, a topological manifold is just a topological space which satisfies certain properties. So, an atlas doesn't actually have anything to do with what a topological manifold is (an atlas just happens to exist on any topological manifold). Two manifolds are equal iff they are equal as topological spaces.
That said, no one actually cares about equality of manifolds. What people actually care about is whether two manifolds are homeomorphic (or more specifically, whether specific maps between them are homeomorphisms). In other words, the "naive equivalence" you are asking about is not important for any applications. As a result, it's perfectly fine to use a definition as you propose in Question 3, where an atlas is part of what a manifold is. This will change what equality of manifolds means (i.e., "naive equivalence") but will not change the notion of equivalence that actually matters, which is homeomorphism.
In the language of category theory, you can define a category $Man$ whose objects are topological manifolds (according to your original definition) and whose maps are continuous maps. You can also define a category $Man'$ whose objects are topological manifolds together with an atlas and whose maps are continuous maps. There is a forgetful functor $F:Man'\to Man$ which forgets the atlas. This functor is not an isomorphism of categories, but it is an equivalence of categories, which is good enough for everything people ever want to do with manifolds.
As a final remark, atlases are pretty irrelevant to the study of topological manifolds. The reason atlases are important is to define smooth manifolds, which impose some additional conditions on what kind of atlases are allowed. A smoooth manifold cannot be defined as just a topological space, but instead must be defined as a topological space together with an atlas satisfying certain assumptions (or a topological space together with some other additional structure equivalent to an atlas).
For smooth manifolds, although an atlas must be included in the definition, there is still an issue similar to your Questions 2 and 3. Namely, multiple different atlases can give "the same" smooth manifold, in the sense that the identity map is a diffeomorphism. This means that if you define a smooth manifold as a triple $(M,T,A)$ where $(M,T)$ is a topological space and $A$ is a smooth atlas on $(M,T)$, then the "naive equivalence" is not the equivalence you actually care about, similar to if you used the definition for topological manifolds you proposed in Question 3.
To avoid this, many authors instead define a smooth manifold as a triple $(M,T,A)$ where $(M,T)$ is a topological space and $A$ is a maximal smooth atlas on $(M,T)$ (or alternatively, $A$ is an equivalence class of smooth atlases on $(M,T)$). This makes the choice of $A$ unique, in the sense that if $(M,T,A)$ and $(M,T,A')$ are smooth manifolds such that the identity map $M\to M$ is a diffeomorphism between them, then $A=A'$. As with topological manifolds, though, it doesn't really matter whether you use this definition or the previous one, since all that changes is what it means for two smooth manifolds to literally be equal and that's not what we actually care about.
Best Answer
The transition map: $$ (f_\alpha\times y_i)\circ (f_\beta\times y_j)^{-1}=(f_\alpha\circ f_\beta^{-1})\times (y_i\circ y_j^{-1}) $$ Since $f_\alpha\circ f_\beta^{-1}$ and $y_i\circ y_j^{-1}$ are both smooth, the product is also smooth, since coordinates of $M$ and $N$ don't interact with each other (their partial derivatives should be consistently $0$, and therefore smooth). Explicitly, $$ (f_\alpha\circ f_\beta^{-1})\times (y_i\circ y_j^{-1}):(x^1,\dots,x^m;a^1,\dots,a^n)\mapsto(z^1,\dots,z^m;b^1,\dots,b^n) $$ From smoothness of $f_\alpha\circ f_\beta^{-1}$ and $y_i\circ y_j^{-1}$, $\partial z^i/\partial x^j$ and $\partial b^i/\partial a^j$ and higher order partial derivatives are all smooth, and the cross terms $\partial b^i/\partial x^j,\partial z^i/\partial a^j$ and the higher order partial derivatives are all $0$ and is therefore also smooth.