Let's take the following definitions.
Definition 1. An Ehresmann connection is a smooth assignment of the complements of the vertical subspaces of the tangent spaces of the total space of the bundle. These complement subspaces are called vertical subspaces.
Definition 2. The horizontal lift of a differentiable curve $C$ in the base space is a curve $D$ in the total space having the following properties
- $\pi(D)=C$
- The tangent vectors of $D$ are horizontal at every point.
Definition 3. An Ehresmann connection on a smooth fiber bundle is curvature free if any horizontal lift of any closed differentiable curve in the base space is also closed. The connections which are not curvature free arre called curved.
Let $M$ be a simply connected 1-dimensional smooth manifold, and $\pi:E\to M$ a smooth fiber bundle.
In the special case when $E=\mathbb R^2$ , $M=\mathbb R$, and $\pi:(x,y)\mapsto (x,0)$, it is fairly obvious, that the horizontal lifts of every closed curve in $M$ are closed. But is this true also, when the fibers are not simply connected, or have more than 1-dimensions?
The topology of the fiber plays essentially no role here - the fact that the base is $\mathbb R$ immediately rules out anything interesting happening.
First, the fact that the base is one-dimensional tells us that the connection is flat, meaning that the lift of any null-homotopic closed curve is closed. The intuition here is pretty simple: such a curve in the base has to backtrack over itself perfectly in order to close, and thus the "vertical motion" of the lift is perfectly reversed. If you know some standard theorems, the neat proof is probably to note that the curvature form $R(X,Y) = [X_H, Y_H]_V$ must vanish since the horiozontal bundle is one-dimensional.
Secondly, the fact the base is simply-connected (in addition to being flat) tells us that the connection has zero holonomy, and thus is "curvature-free" in your parlance. This is super easy: simply-connected means exactly that every curve is null-homotopic.