In my other answer, I showed that the only feasible definition of those you give is the first one, with $\kappa\leq N$. For example, one might hope to construct a countable approximation $\mu$ to 1-D Lebesgue measure $\lambda$ and take $\kappa=1$, so that $$|\mu(B(x,r))-\lambda(B(x,r))|<O(r)$$ for small $r$. (Note that this does not say much: since $\lambda(B(x,r))=2r$, this is equivalent to requiring $$0\leq\mu(B(x,r))\leq O(r)$$ Nevertheless…) I am unable to construct such a measure.
In fact, that example must contain essentially all the difficulty: let $X$ have c.d.f. $F$. Then $F(X)$ is a random variable uniformly distributed on $[0,1]$. If $\mu$ is approximates Lebesgue measure, then the pull back of $\mu$ via $F$ must approximate the law of $X$.
Below is the closest I have come to constructing such a measure.
Let $$D_n=\left\{\frac{a}{2^n}:|a|<2^n,2\nmid a\right\}$$ denote the dyadic rationals of level $n$ in $(-1,1)$ and $D=\bigcup_{n=0}^{\infty}{D_n}$. Then $|D_n|=2^n$ and $D$ is countable.
We let $\mu({x})=2^{-2n-1}$ for $x\in D_n$; this defines a probability measure on the power set of $D$.
We are interested in counting the intersection of $D_n$ with balls, so fix $n$ and choose a point $x\notin D$ and a radius $2^{-k}$. Note that the points of $D_n$ are spaced $\frac{1}{2^{n-1}}$ apart and intercalate between $D_{n-1}$ exactly. Thus the elements of $\bigcup_{n<k}{D_n}$ are spaced distance $2^{1-k}$ apart. Since $x\notin D_k$, we cannot have $\partial B(x,2^{-k})\subseteq D_{k-1}\sqcup D_{k-2}$; thus $$\left|B(x,2^{-k})\cap\bigcup_{n<k}{D_n}\right|=1\tag{1}$$ Likewise, if $k\leq n$, then $$|B(x,2^{-k})\cap D_n|=2^{n-k}\tag{2}$$
We can improve on (1) if we allow ourselves access to the binary representation of $x$. First, write $x=(-1)^s\sum_{j=0}^{\infty}{\frac{a_j}{2^j}}$ where each $a_j\in\{0,1\}$. Second, round $x$ to $(k-1)$-many digits, producing $y$; third, let $\delta_k(x)$ be one plus the index of the last "$1$" appearing in $\{a_j\}_{j<k}$. (If $y=0$, then $\delta_k=0$.)
By definition, $y\in D_{\delta_k(x)}$. In addition, we have constructed $y$ so that $|x-y|<\frac{1}{2^k}$ ($x\notin D$ rules out equality); thus $y\in B(x,2^{-k})$. So if $k<n$, we have
$$\left|B(x,2^{-k})\cap D_n\right|=\begin{cases}
1 & n=\delta_k(x) \\
0 & n\neq\delta_k(x)
\end{cases} \tag{3}$$
Finally, we can compute: \begin{align*}
\mu(B(x,2^{-k}))&=\frac{1}{2^{2\delta_k(x)+1}}+\sum_{n=k}^{\infty}{\frac{2^{n-k}}{2^{2n+1}}} \\
&=\frac{1}{2^{2\delta_k(x)+1}}+2^{-k-1}\sum_{n=k}^{\infty}{2^{-n}} \\
&=\frac{1}{2^{2\delta_k(x)+1}}+\frac{1}{2^{2k}}
\end{align*}
Writing $r=2^{-k}$, we can summarize this as: $$\mu(B(x,r))=r^2+\frac{1}{2^{2\delta_k(x)+1}}$$ $r^2=O(r)$ for small $r$, so that's good; what about the other term? Well, since $x\notin D$, we know that $\lim_{k\to\infty}{\delta_k}=\infty$. But we do not have any bounds on $\delta_k$ uniform in $x$, and in fact cannot. There are transcendental numbers with arbitrarily large gaps between "$1$"s in their digit sequence; $x$ might be one such. (We can find a bound that holds for $\lambda$-a.e. $x$: \begin{align*}
\lambda\left(\liminf_{k\to\infty}{\left\{x:\delta_k(x)>\frac{k}{\log{(k)}}\right\}}\right)&=\liminf_{k\to\infty}{\lambda\left(\left\{x:\delta_k(x)>\frac{k}{\log{(k)}}\right\}\right)} \tag{*} \\
&=1-\limsup_{k\to\infty}{\lambda\left(\left\{x:\delta_k(x)\leq\frac{k}{\log{(k)}}\right\}\right)} \\
&=1-\limsup_{k\to\infty}{\sum_{d\in\{0,1\}^{\frac{k}{\log{(k)}}}}{2^{-k}}} \tag{†} \\
&=1-\limsup_{k\to\infty}{2^{\frac{k}{\log{(k)}}-k}} \\
&=0
\end{align*} where (*) follows from the Dominated Convergence Theorem and (†) from the characterization of $\left\{x:\delta_k(x)\leq\frac{k}{\log{(k)}}\right\}$ as "numbers with all binary digits from index $\frac{k}{\log{(k)}}$ to $k$ equal to 0.")
However, Hurwitz's theorem tells us that any irrational $x$ can be approximated by rationals very closely: there are infinitely many $p,q$ such that $$\left|x-\frac{p}{q}\right|<\frac{1}{\sqrt{5}q^2}$$ This suggests that, if we place atoms at not only the dyadic rationals, but all rationals, then we might be able to force $\delta$ to grow at a reasonable rate. But when I try to write down such a measure, I find calculations of the sort (1-3) are beyond my ability. Perhaps you, dear reader, can complete them, bound $\delta$, and solve the problem.