sphere bundles isomorphic to vector bundle. Bredon claims that sphere bundles in certain cases are isomorphic to vector bundles. For example he says just replace $S^{n-1}$ with $R^n$. But for example the circle and the plane are not even isomorphic. How can we talk about bundle isomorphism when even the fibers are not isomorphic?
"A disk or sphere bundle gives rise to a vector bundle with the orthogonal group as the structure group just by replacing the fiber $D^n$ or $S^{n-1}$ with $R^n$ and using the same change of coordinate function".
We cannot just "replace" the fibers with something not isomorphic can we?

Apparently, in this context you have a sphere bundle say $M$ with the orthogonal group as structure group. This is a very special case of a sphere bundle. This means that you have a family of trivialization $\phi _i$ of your bundel $F$ over open sets $U_i$, say $\phi _i : U_i\times S \to M$ such that the change of chart over $U_i\cap U_j$ is of the form $\phi _{i,j} (u, v)= O_{i,j}(u).v$ where $O_{i,j}: U_i\cap U_j \to O(n)$ is a certain (continuous or smooth) function. Bredon says that you can use the same functions to define an $\bf R^n$ bundle $E$. First defined  its trivialization over $U_i$, say $E_i$ by the formula $\Phi _i : U_i\times {\bf R}^n  = E _i$. Then $E$ is the union of the $E_i=U_i\times {\bf R}^n$ glued over $U_i \cap U_j$ by the map $\phi _{i,j} (u, v)= O_{i,j}(u).v$.
Another approach is through the theory of principal bundles. To a sphere bundle with $O(n)$ as structure group you can associate its frame bundle $P$, a principal $O(n)$ bundle, which is the bundle whose fiber at a point $p$ is the set of isometries between the fiber over the point and the standard $n$ sphere. Once this frame bundle is defined, you can construct its associate euclidan bundle by the formula $E= P\times _{O(n)} {\bf R}^n$ 
