If I have a loop $\gamma(t)$ in a manifold $M$, $\gamma(0) = \gamma(1)$, with a homotopy $\gamma_s(t)$ such that $\gamma_0(t) = \gamma(t)$ and $\gamma_1(t)$ the constant curve $\gamma(0)$, consider the set of points
$$S = \{ p \in M | \exists s,t \in I,\ p \in \gamma_s(t) \}$$
If I consider a tubular neighbourhood $T(S)$ around that set (by which I mean that for every point $p \in S$, I consider an open ball around $p$), with the topology from $M$, is this resulting set homeomorphic to $\Bbb R^n$? In particular, is there a coordinate chart $\phi : T(S) \to \Bbb R^n$?
Assuming I'm understanding your question correctly, the tubular neighborhood need not even be contractible, let alone homeomorphic to Euclidean space. Take $M = \mathbb{C} \cong \mathbb{R}^2$ the complex plane, and let $$\gamma_s(t) = \exp(2\pi i (1-s) \cdot 2(\lvert t - 0.5 \rvert - 0.5)).$$ Essentially $\gamma = \gamma_0$ is the path winds along the unit circle once, and then backtracks, and the homotopy keeps track of how far around the circle the path goes. The image of $\gamma$ is precisely the unit circle, and a sufficiently small tubular neighborhood of it looks like an annulus, which is not simply-connected.