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
It's not the covering that's the (potential) problem, it's the compatibility, but actually this won't be an issue for topological atlases.
Let $U_{\alpha_1\alpha_2} = U_{\alpha_1} \cap U_{\alpha_2}$. For the two charts $\phi_{\alpha_1}, \phi_{\alpha_2}$ to be compatible, you need $$\tau_{1,2}=\phi_{\alpha_2} \circ \phi_{\alpha_1}^{-1}: \phi_{\alpha_1}(U_{\alpha_1\alpha_2}) \to \phi_{\alpha_2}(U_{\alpha_1\alpha_2})$$ to be continuous (in fact a homeomorphism, but swapping the indices gives the continuous inverse). But this is automatic, since both $\phi_{\alpha_1}$ and $\phi_{\alpha_2}$ are generally required to be homeomorphisms.
The trouble comes in when you want to look at atlases with more structure. For example, a "smooth atlas" is one where you require the transition maps $\tau_{i,j}$ to be smooth, not just continuous. Then there is the possiblity that two charts will be incompatible.