Yes, the condition above implies a local isomorphism from the section of a neighborhood of x in E to the bundle X×Rn. This is because the measurable maps πi are defined to be isomorphisms between the sections of E and the bundle X×Rn in the neighborhoods Yi.
However, this definition alone does not imply that E is a fiber bundle over X. A fiber bundle is a special type of bundle that has additional structure, such as a continuous surjective map from the bundle to the base space, a continuous right inverse map, and a smooth structure on the fibers. In order to show that E is a fiber bundle, it would be necessary to show that these additional conditions are satisfied.
The definition you provided for a measurable bundle does not guarantee the local product structure. Specifically, the condition you provided only states that there exists a countable cover of X by measurable sets $Y_i$ and measurable maps $\pi_i: X \times \mathbb{R}^n \to X$ such that the inverse image of $Y_i$ under $\pi_i$ is equal to $Y_i \times \mathbb{R}^n$. This does not necessarily imply that there exists an open set $U$ in X such that the inverse image of $U$ under $\pi$ is isomorphic to $U \times \mathbb{R}^n$.
However, if you further assume that the measurable bundle is a fiber bundle, then it will have the local product structure. A fiber bundle is a type of bundle that has a locally trivializable projection map and a fiber, which is a topological space that is homeomorphic to the inverse image of a point under the projection map. In the case of a measurable fiber bundle, the projection map and the fiber are required to be measurable as well. The local trivializability condition guarantees that there exists an open set $U$ in X such that the inverse image of $U$ under the projection map is isomorphic to $U \times F$, where $F$ is the fiber. This implies that the local product structure holds for measurable fiber bundles.
