The Cartesian product of two topological spaces \(M\)and \(N\) can be thought of as the set of all ordered pairs selected from \(M\) and \(N\). (There's more to the definition than this, but this is what's pertinent for our purposes.) For example, the Cartesian product of two open/closed intervals is an open/closed rectangle. This means that if you're going to tell me that AdS_{n} has the same topology as \(\mathbb{R}^n \times S^m\), I should always be able to pick a point from the Euclidean space and a point from the m-sphere, and you can show me the point in AdS_{n} that corresponds to that ordered pair.

With that in mind: if you write your constraint as

\((x^0)^2 + (x^1)^2 = \alpha^2 + \vec{x}^2\),

then it's not too hard to see that this defines a circle in the \(x^0\text{-}x^1\) plane *for any value of *\(\vec{x}\), with radius \(\sqrt{\alpha^2 + \vec{x}^2}\). This means that if I pick a point on the circle \(S^1\), and pick a point in the Euclidean space \(\mathbb{R}^{n-1}\), then there will always be a unique point in AdS_{n} corresponding to that choice.

On the other hand, if you write

\(\vec{x}^2 = (x^0)^2 + (x^1)^2 - \alpha^2\),

then this does not define an (n-2)-sphere for all possible choices of \(x^0\) and \(x^1\), since the right-hand side can become negative. Thus, your proposed map from ordered pairs in \(\mathbb{R}^2\) and \(S^{n-2}\) to AdS_{n} doesn't work, since I can choose points from the base spaces that have no corresponding point in AdS_{n} (namely, any point from \(\mathbb{R}^2\) that has \((x^0)^2 + (x^1)^2 < \alpha^2\), and any point at all from \(S^{n-2}\).)

Of course, this flaw in your second argument doesn't prove on its own that AdS_{n} isn't homeomorphic to \(\mathbb{R}^2 \times S^{n-2}\)—merely that this map isn't a homeomorphism. But since we already found another argument that AdS_{n} was homeomorphic to \(\mathbb{R}^{n-1} \times S^1\), and since these two product spaces aren't homeomorphic to each other (they're topologically distinct, so they can't be), we can conclude that AdS_{n} isn't homeomorphic to \(\mathbb{R}^2 \times S^{n-2}\). (Unless, of course, \(n = 3\).)