I am working with the following definition of fiber bundle:
Let $M$ and $F$ be topological spaces. A fiber bundle over $M$ with model fiber $F$ is a topological space $E$ together with a surjective continuous map $\pi \, \colon E \to M$ satisfying the following property: for each $p \in M$, there exists a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $\Phi \, \colon \pi^{-1}(U) \to U \times F$ such that $\pi_{1} \circ \Phi = \pi$, where $\pi_{1} \, \colon U \times F \to U$ is the projection on the first factor.
Now, let $(E,\pi)$ be a fiber bundle. Clearly, denoting $\pi^{-1}(\{p\})$ by $E_{p}$, one has $$ E = \bigcup_{p \in M} E_{p}, $$ and each fiber $E_{p}$ is homeomorphic to $F$. I claim that one can always write $$ E_{p} = \{p\} \times F_{p}, $$ for some $F_{p} \cong F$, i.e., the total space is the disjoint union of the collection $\{F_{p}\,\colon \,p \in M\}$. Am I right?
You can't really write $E_p = \{ p \} \times F_p$ because $E_p$ is a subset of $E$ and there's no guarantee that it will be of the form $\{ p \} \times F_p$ for some $F_p$ simply because it doesn't have to look like a product (a set). It is true that often one constructs a bundle over $M$ by defining $E$ to be $E = \bigcup_{p \in M} \{ p \} \times F_p$ where each $F_p$ is diffeomorphic to $F$ and then topologizing $E$ in an appropriate way but this is just one way of constructing a fiber bundle.
For an example, think of the map $\pi \colon S^1 \rightarrow S^1$ given by $\pi(z) = z^2$. This is a covering map where each fiber has two points so you can think of it as a fiber bundle with typical fiber (say) $F = \{ 7, 9 \}$. The fiber $\pi^{-1}(1)$ is the set $\{-1, 1 \}$ but it makes no sense to say $\{ -1, 1 \} = \{1 \} \times \{ a, b \}$.