Leray-Hirsch statement Theorem explanation This is on p50 Bott & Tu's Differential Forms

(Leray-Hirsch) Let $E$ be a fiber bundle over $M$ with fiber $F$. Suppose $M$ has a finite good cover. If there are global cohomology classes $e_1, \ldots, e_r$ on $E$ which when restricted to each fiber freely generate the cohomology of the fiber, then $H^*(E)$ is a free module over $H^*(M)$. 
  $$H^*(E) \cong H^*(M) \otimes \Bbb R \{e_1, \ldots, e_r \} \cong H^*(M) \otimes H^*(F)$$ 

I am unsure - but does the hypothesis implies $E \cong M \times F$ as fiber bundles? 

Here we have fiber space $F = \Bbb R^n$, in which our topological group $G$ that acts freely on $F$ is $GL_n(\Bbb R)$. 
Then there is a map 
$$ M \times F \rightarrow E, \quad (m,v) \mapsto \sum v_i e_i (m)$$ 
where $\sum v_i e_i(m)$ is in the fiber of $m$. The map is an isomorphism as restricted to each map, it is an $\Bbb R$-linear continuous isomoprhism. And as a general result of topological vector bundles, this is a homeomorphism. 
 A: This is not true. 
Firstly, a fibre bundle is a much more general object than a vector bundle, which is just a fibre bundle with typical fibre a vector space, that satisfies some additional conditions.
Note that your example is incorrect. In general there is no such map, since you are relying on the existence of a collection $(e_i)_{i=1,\dots,n}$ of global non-vanishing sections $M\rightarrow E$. Such a collection exists if and only if the bundle is trivial. If the bundle is trivial, then the map you write down is indeed a global trivialisation, providing the bundle isomorphism $E\cong M\times \mathbb{R}^n$. Conversely, if $\varphi:M\times F\xrightarrow{\cong} E$ is a global trivialisation, then the $n$ maps $e_i:M\rightarrow \mathbb{R}^n$, $m\mapsto \varphi(m,v_i)$, where $v_i\in\mathbb{R}^n$ is the $i^{th}$ standard basis vector, are a collection of $n$ non-vanishing global sections.
In general there do exist non-trivial vector bundles. For instance, the Möbius bundle $E\rightarrow S^1$. You can check quite easily that there are no sections $S^1\rightarrow E$ that do no vanish at at least one point of $S^1$.
As an example of a fibre bundle meeting the assumptions of the Leray-Hirsch theorem you state, consider the bundle 
$$S^2\xrightarrow{i}\mathbb{C}P^3\xrightarrow{p} S^4.$$
Here you have
$$H^*\mathbb{C}P^3\cong \mathbb{Z}[x_2]/(x_2^4)\cong \{1,x_2\}\otimes H^*S^4\cong H^*S^2\otimes H^*S^4$$
where $i^*x_2=s_2\in H^2S^2$ is a generator, and for a generator $s_4\in H^4S^4$, it holds that $p^*s_4=x_2^2\in H^4\mathbb{C}P^3$.
