Let $E$ be an oriented vector bundle over $B$, a CW complex, with fiber of dimension $n$. Let $E_0$ be $E - B\times 0$. The main theorem Hatcher uses to prove the thom isomorphism theorem is that the restriction map $H^n(E,E_0) \to H^n(R^n,R^n-0)$ is an isomorphism.
But here is a counter example: Let $E=S^2 \times R^2$. Being a trivial bundle it is oriented. We have $H^2(E,E_0)=H^2(S^2 \times (D^2, S^1))=\tilde H^2(S^2 \times (D^2, S^1))=\tilde H^2(S^2 \times S^2)=H^2(S^2 \times S^2) =\mathbb{Z} \oplus \mathbb{Z}$.
But $H^2(R^2,R^2-0)=Z$. So there cannot be an isomorphism between $H^2(R^2,R^2-0)$ and $H^2(E,E_0)$.
What is wrong?
Your mistake is in the equality $\tilde{H}^2(S^2 \times (D^2, S^1)) = \tilde{H}^2(S^2 \times S^2)$ you wrote. It might be a bit easier to see why this is false in one dimension lower. Consider the trivial bundle $S^1 \times D^1 \to S^1$ instead. In the proof you would consider $(S^1 \times D^1, S^1 \times S^0)$, and it might be tempting to write $H^1(S^1 \times (D^1, S^0)) = H^1(S^1 \times S^1)$.
But this is false! When you look at the quotient $(S^1 \times D^1) / (S^1 \times S^0)$ (which you implicitly do), this is not a torus. You collapse all the boundary to one point, and so you get a torus with the whole inner circle collapsed, or equivalently a 2-sphere with two points identified. And the first cohomology group of this is $\mathbb{Z}$ as expected, in fact it's homotopy equivalent to $S^2 \vee S^1$.
More generally the problem is that $(X \times Y) / (X \times Z)$ is not $X \times (Y/Z)$.