Consider some kind of bundles, for instance vector bundles or fibre bundles with a certain structure group, such that there are bundle morphisms and pull-backs. 


Let then $F(X)$ denote the isomorphism classes of bundles over $X$, this is a contravariant functor by virtue of the pull-back construction. 
Assume further that $F(.)$ satisfies the conditions of Brown's representabiliy theorem, which is usually the case in such a setting. 


So we find a space $B$ such that there is a natural isomorphism $F(X) \cong [X,B]$.
Let $ \gamma$ denote the bundle, which corresponds to [id\_B] under this isomorphism.



> 
> Is $\gamma$ then a universal bundle?
> 
> 
> 


I can prove that for every bundle there is a morphism to $\gamma$ and that the base map of this morphism is unique up to homotopy. It is however not clear to me, why the morphism itself is unique up to homotopy. 


If we consider for instance orientable or oriented vector bundles, then $\gamma$ has a self-isomorphism induced by sending every vector to its negative and in odd dimension this is not orientation preserving, so the automorphism can not be homotopic to the identity as a morphism. Therefore $\gamma$ cannot be a universal bundle. 



Is there always a universal bundle over a classifying space?
There seems to be a little confusion here. Let me try to be elementary. For any type of bundle (or fibration)
closed under pullbacks, a bundle $\gamma \colon E\to B$ of the specified type is defined to be universal if pullback 
of $\gamma$ along maps $f\colon X \to B$ induces a natural isomorphism $[X,B]\to F(X)$, where $F(X)$ is the
set of isomorphism classes of bundles of the speciified type, and naturality refers to maps that preserve whatever
structure that class of bundles has. That is the (or a) standard definition in algebraic
topology; here one might want to restrict the $X$ to have the homotopy types of CW complexes if one has
classical insecurities. Then, by definition, $\gamma$ is a universal bundle and is the bundle which corresponds 
to $[id\_B]$ under the isomorphism. When one constructs $\gamma$ by Brown representability, one is 
representing the functor $F$, so getting the required bundle $\gamma$. The fallacy in the last paragraph of
the question is that one cannot take orientable bundles but one can take oriented bundles, so that a map
of bundles must preserve the orientation. Then the supposed contradiction disappears.


I don't want to go into details on the definitions in <https://arxiv.org/abs/math/0105047>, but they are fairly
non-standard from the usual point of view of algebraic topology. The idea of 2.2 seems to be that there should be no
obstructions to extending bundles, as should follow or be equivalent to the contractibility of the total space of 
$\gamma\_{TOP}^n$. There is no uniqueness claim in the extendability condition, even up to bundle homotopy.
Then 2.3 gives the uniqueness of the classifying map (not classifying morphism, which is a morphism of bundles,
not of base spaces). In 2.4 he is saying explicitly that classifying maps (base space level) are unique up to
homotopy but that classifying morphisms (bundle level) need not be unique up to homotopy, as was illustrated 
(or meant to be illustrated) by the example of trivial bundles. Judging by the number of typos, that paper seems to have been 
written in haste, but it doesn't seem obviously wrong.


Incidentally, although the ordinary cohomology of $BTop$, the colimit of the $BTop(n)$, is well understood, the mod 
$p$ cohomologies of the $Top(n)$ are still not fully computed as far as I know. 


