Consider a compact Riemannian manifold $(M, g)$. Choose a ball $B(p, r)$ inside $M$, and a quasi-isometric ball $B(q, s)$ in $\mathbb{R}^n$, in the image of a coordinate chart containing $B(p, r)$ (in this context, quasi-isometric means that $c_1 g_{\mathbb{R}^n} \leq g \leq c_2 g_{\mathbb{R}^n}$ on $B(p, r)$ and $B(q, s)$. Now, consider a compact set $K \subset B(p, r)$ and its image $K'$ inside $B(q, s)$ under the coordinate chart.

It seems intuitively that the probability of a Brownian particle (corresponding to the Brownian motion on $M$) starting at $p$ and hitting $K$ within small time $t$ is comparable to the probability of a Brownian particle (corresponding to the Brownian motion in $\mathbb{R}^n$) starting at $q$ hitting $K'$ within (the same) small time $t$. How can one prove this?

To rephrase, is there a precise estimate to fit the following heuristic: if the metric is locally slightly perturbed, then the Brownian motion for small times is also slightly perturbed?

Edit: An issue seems to be that there needs to be a global argument somewhere. For example, one somehow needs to say that the contribution from paths which go far out and then come back again in short time is negligible in both cases. This seems very believable intuitively, as a Brownian motion effectively travels a distance $\sim t^2$ in time $\sim t$, but again, I am uncertain how to formalize this.