As a first observation, if $(E,\nabla)$ has this Rolle property then if $i: S^1 \to M$ is any embbeding, $(i^*E,i^*\nabla)$ has this property as well (this is just a tubular neighborhood + cutoff argument). In particular we need to understand what possible such vector bundles with connection appear over $S^1$. It is fairly clear that the only possibilities are line bundles. If $i^*E$ is the Mobius bundle, then any connection has this property. If it is the trivial bundle, then the connection must have trivial holonomy (in other words there must be a trivialization for which the connection $i^*\nabla$ is the canonical one associated to that trivialization). In particular the connection must be flat.

For $\dim M > 2$ this classifies all such bundles rather satisfactorily: they are precisely the line bundles with a metric, equipped with the unique connection that preserves this metric. This is because any embedded circle in $\dim \geq 3$ admits a family of embeddings $i_n:S^1 \to M$ converging to the immersion of multiplicity two $z \mapsto i(z^2)$. The holonomy along these converges to the holonomy along the multiplicity two embedding, so the holonomy along $i$ must be $-1$ (a priori it could have been any negative scaling).

In dimension two, the picture can be more complicated. The problem arises if $i$ embeds $S^1$ with trivial normal bundle, but $i^*E$ is nontrivial. In this case, the holonomy may be a negative number $\neq -1$ and thus the connection may not be metric compatible. For example, consider the flat line bundle on $\mathbb{T}^2$ which has holonomy $-\lambda$ along one generator of the fundamental group, and trivial holonomy along another. Note on the other hand that $\mathbb{R}P^2$ has no such problems as an embedding is either trivial or has nontrivial normal bundle.

As a first observation, if $(E,\nabla)$ has this Rolle property then if $i: S^1 \to M$ is any embbeding, $(i^*E,i^*\nabla)$ has this property as well (this is just a tubular neighborhood + cutoff argument). In particular we need to understand what possible such vector bundles with connection appear over $S^1$. It is fairly clear that the only possibilities are line bundles. If $i^*E$ is the Mobius bundle, then any connection has this property. If it is the trivial bundle, then the connection must have trivial holonomy (in other words there must be a trivialization for which the connection $i^*\nabla$ is the canonical one associated to that trivialization). In particular the connection must be flat.

For $\dim M > 2$ this classifies all such bundles rather satisfactorily: they are precisely the line bundles with a metric, equipped with the unique connection that preserves this metric. This is because any embedded circle in $\dim \geq 3$ admits a family of embeddings $i_n:S^1 \to M$ converging to the immersion of multiplicity two $z \mapsto i(z^2)$. The holonomy along these converges to the holonomy along the multiplicity two embedding, so the holonomy along $i$ must be $-1$ (a priori it could have been any negative scaling).

In dimension two, the above argument breaks down, and indeed there are line bundles on $S^1 \times \mathbb{R}$ that have holonomy a negative number $\neq 1$. However, there is clearly some reasonable casework that could be done: on $\mathbb R P^2$ all line bundles satisfying the condition are metric compatible: all homotopically nontrivial embedded circles have nontrivial normal bundle, so we may find embeddings $i_n$ converging to $i$ with multiplicity two. In the torus, we simply see that there are no holonomy maps $\mathbb{Z}^2 \to \mathbb{R}^{\times}$ which take on values besides $\{ \pm 1\}$ but also take on negative values or $1$ at every class representable by an embedding (namely if some class takes on a value beside $\pm 1$, some class representing an embedding takes on a positive value not equal to $1$), so again all line bundles satisfying our condition are metric compatible. (this section was edited)