It seems such a collection of balls exists, for any smooth Riemannian manifold.

First of all, for each $p\in M$ let $B_p$ be a precompact normal ball centered at $p$ so small that:

 1. $\mu(\partial B_p)=0$ (we can do this because $\{r>0;\mu(\partial B(p,r))>0\}$ is countable).
 2. The injectivity radius of points in $B_p$ is bounded below by some constant $\varepsilon>0$. We can prove that a small enough ball achieves this by changing the metric of $M$ far from $p$ so that $M$ becomes complete, and then using that in a complete Riemannian manifold the [injectivity radius is continuous](https://mathoverflow.net/questions/53381/the-continuity-of-injectivity-radius#:~:text=Berger%2C%20I%20learned%20that%20the,curved%20point%20on%20a%20paraboloid.).
 3. Sectional curvatures inside $B_p$ are in some compact interval $[a,b]$.

**Claim 1:** If $\delta>0$ is small enough, then for any $q\in B_p$ we have $\frac{\mu(B(q,\delta))}{\mu(B(q,2\delta))}>2^{-n-1}$, where $n$ is the dimension of $M$.

Proof: This constant $\delta$ will be $<\frac{\varepsilon}{2}$, and to prove it exists, first note that due to Theorem 3.23 in [1] we have $\frac{\mu(B(q,\delta))}{\mu(B(q,2\delta))}\geq\frac{V_b(\delta)}{V_a(2\delta)}$, where $V_\kappa(r)$ is the volume of the ball of radius $r$, in the $n$-dimensional space of constant sectional curvature $\kappa$. Note that for any $x>0$ we have $V_\kappa(xr)=x^nV_{\kappa x}(r)$, because the space of constant curvature $r$ is obtained from multiplying the metric of the space of constant curvature $xr$ by $x^2$. So when $\delta\to0$, we have $\frac{V_b(\delta)}{V_a(2\delta)}=\frac{V_{\delta b}(1)}{V_{\delta a}(2)}\to\frac{V_0(1)}{V_0(2)}=2^{-n}$, which proves that the constant $\delta$ exists. $\square$

**Claim 2:** For any open subset $A$ of $B_p$ there is a finite set of balls $B_1,\dots,B_m$ contained in $A$ such that $\mu(\cup_{i=1}^m B_i)>\frac{1}{2^{n+2}}\mu(A)$.

Proof: Let $\delta$ be so small that it satisfies the previous claim and if $B:=\{x\in A;d(x,M\setminus A)>\delta\}$, then $\mu(B)>\frac{1}{2}\mu(A)$. Now consider a maximal $2\delta$-separated set $\{x_1,\dots,x_m\}$ in $A$, and let $B_i:=B(x_i,\delta)$. These balls are disjoint, and $\sum_i\mu(B_i)>\frac{1}{2^{-n-1}}\sum_i\mu(B(x_i,2\delta))\geq\frac{1}{2^{-n-1}}\mu(B)\geq\frac{1}{2^{n+2}}\mu(A)$, because the balls $B(x_i,2\delta)$ cover $B$. $\square$

We can also ensure in that proof that the boundaries of the balls $B_i$ of the previous claims have measure $0$: if not, note that for each $q\in M$, the set $\{r>0;\mu(\delta B(q,r))>0\}$ is countable, so we can reduce the radii of the balls just a little bit so that the sum of their volumes is still $>\frac{1}{2^{-n-1}}\mu(A)$.


**Claim 3:** We can cover any open set $X\subseteq B_p$ up to measure $0$ by a disjoint collection of balls contained in $X$.

Proof: Take $A=X$ in the previous claim, and find balls $B_{0,1},\dots,B_{0,n_0}$ with boundary of measure $0$ such that $\mu(\cup_{i=1}^m B_i)>\frac{1}{2^{n+2}}\mu(X)$. Now let $X_1=X\setminus\cup_i\overline{B_{0,i}}$, so that $\mu(X_1)\leq(1-\frac{1}{2^{n-1}})\mu(X)$. Applying the same to $X_1$ we can remove from it finitely many balls $B_{1,1},\dots,B_{1,n_1}$ to obtain some open $X_2$ with $\mu(X_2)\leq(1-\frac{1}{2^{n-1}})\mu(X_1)$. Repeating this step to obtain spaces $X_n$ for each $n$, we get that the balls $\{B_{i,j}\}_{i=1,\dots,n_j}$ are pairwise disjoint, and $\mu(X\setminus\bigcup_{i,j}B_{i,j})=\lim_{m\to\infty}\mu(X_m)\leq \lim_{m\to\infty}(1-\frac{1}{2^{-n-1}})^m\mu(X)=0$. $\square$

**Claim 4:** We can cover $M$ up to measure $0$ using a countable collection of disjoint compact balls.

Proof: Consider the collection of balls $\{B_p;p\in M\}$. As $M$ is second countable, we can find a countable subcover of $M$, $(B_n)_{n\in\mathbb{N}}$. Moreover, for each $n$, we can cover $B_n\setminus\bigcup_{i=1}^{n-1}\overline{B_i}$ up to measure $0$ with a countable collection of disjoint compact balls. The union of these countable collections of balls covers $B_n$ up to measure $0$ for all $n$, thus it covers all $X$ up to measure $0$.

[1] Cornelia Druţu, Michael Kapovich, \textit{Geometric Group Theory}, Colloquium Publications. Volume: 63; 2018.