Are you allowing $b$ to depend on the manifold (as it appears to me from your statement)? In that case, Besicovitch is overkill, and this statement holds in much more generality than Besicovitch. One only needs to know that a compact Riemannian manifold is a doubling metric space. Doubling has many equivalent definitions, but here you can think of it as: any ball of radius $s$ contains at most $C$ $s/2$-separated points, for some fixed constant $C$ independent of $s$.
With that, let $N$ be a maximal $\frac{r}{2}$-separated set in $M$, i.e., a set of points such that any two have distance $\geq \frac{r}{2}$ and such that every point in $M$ is distance $<\frac{r}{2}$ away from one of them. Let $\mathcal{U}$ be the balls of radius $r$ centered on $N$. By construction $\mathcal{U}$ satisfies (ii) immediately. Property (i) follows from the doubling condition: The centers of all balls in $\mathcal{U}$ that touch a fixed ball from $\mathcal{U}$ form an $\frac{r}{2}$-separated set inside a ball of radius $2r$, and the cardinality of such a set is controlled by the doubling constant.
To see that every compact Riemannian manifold $M$ is doubling is easy. Cover $M$ by finitely many charts that are bi-Lipschitz with constant $2$ to subsets of $\mathbb{R}^d$, and the doubling constant can then be estimated based on the number of charts and the doubling constant of $\mathbb{R}^d$. More refined estimates based on curvature are possible too; it depends what you need.