On defining a fibre bundle, it is argued that the projection map $\pi$ requires to be satisfied the condition that there is a homeomorphism $h$ such that the first coordinate coincide.
Definition (Fibre bundle). A fibre bundle structure on a space $E$ is a pair of spaces $F$ and $X$ equipped with a projection map $\pi: E \rightarrow X$ such that there is an open cover $\left\{U_i\right\}$ of $X$ and homeomorphisms $h_i: \pi^{-1}\left(U_i\right) \rightarrow U_i \times F$ coinciding with $\pi$ in the first coordinate. We write: $$ F \longleftrightarrow E \stackrel{\pi}{\longrightarrow} X . $$ In particular, this means that each fibre $F_x=\pi^{-1}(x)$ maps homeomorphically to $\{x\} \times F$. We call $E$ the total space, $X$ the base space, $F$ the fibre and $h_i$ the (local) trivialisations.
But I don't understand how this two spaces $\pi^{-1}\left(U_i\right)$ and $U_i \times F$ are really different?
They are not "really" different in the sense that they are diffeomorphic, meaning "essentially the same".
However, they are not "the same" in the sense that $\pi^{-1}(U_i)\subseteq E$ and $U_i\times F$ is not related to $E$ (Note that $E$ and $F$ are both manifolds).
So the statement that $E$ has a fibre structure means that locally the space $E$ looks like $U_i\times F$, i.e. there are open subsets $\pi^{-1}(U_i)$ that look like/are diffeomorphic to $U_i\times F$.
However, $E$ might not have the global structure $E\cong X\times F$ (not necessarily true). For example, let $E$ be a Moebius strip and let $X$ be a circle on this Moebius strip (going around once). Now for each $x\in X$ draw a line segment on the strip such that the whole strip is nicely divided up into line segments (As in this wikipedia image). Then globally the space $E$ does not look like "a circle"$\times$"a line segment". That would be a cylinder, not a Moebius strip.