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Everywhere i see definition of vector bundle as triple $(E, p, B)$, $B$ and $E$ are manifold and local trivialization condition holds. For example see the definition here. .
Local trivialization gives manifold structure on $E$, then why initial assumption on $E$ that $E$ is manifold is necessary.
You can avoid assuming that $E$ is a smooth manifold, but you must then ensure that the trivializations you've put forth induce a smooth structure on $E$. In other words, if $\varphi\colon p^{-1}(U) \to U \times \mathbb R^k$ and $\psi\colon p^{-1}(V) \to V \times \mathbb R^k$ are to be charts for $E$ then $\psi \circ \varphi^{-1}$ must be a diffeomorphism from $(U \cap V) \times \mathbb R^k$ to itself. This leads to the idea of specifying vector bundles via transition functions.
All you need is that $E$ and $B$ are topological spaces Perhaps a consequence of the local triviality condition is that $E$ is a topological manifold, but I've never seen it presented as an axiom.
Take a look at this definition.
In the link you provide, $E,B$ are only assumed to be topological spaces, and $p:E\to B$ is a continuous surjective map.
If you are interested in smooth vector bundles then assuming that $E$ only is a topological space and $B$ is a smooth manifold is enough, since local trivialization of $p$ will give $E$ the structure of a smooth manifold. But you still need $p$ to be assumed continuous, otherwise you cannot form a smooth atlas.
