This was initially intended to be a reply to geetha290krm's supposed counterexample to (2) (showing it is $\textbf{not}$ in fact a valid counterexample, and that in the context of $X$ being locally compact Hausdorff, (2) will always in fact be true), but was too long for a comment, so I wrote it up as an answer.
Lets work in $\mathbb{R}_{> 1}$, for simplicity (with the Lebesgue measure).
Let $0 \leq \chi_{n} \leq \frac{1}{n}$ be a bump function equal to $1/n$ on $[n + \epsilon_{n}, n+1 - \epsilon_{n}] $, with support contained in $[n + \epsilon_n^2 ,n + 1 - \epsilon_n^2] \subset (n,n+1)$ for each $n$ , where $\frac{1}{4}> \epsilon_{n} > 0$ are chosen sufficiently small so that $\epsilon_{n} \downarrow 0$ as $n \uparrow \infty$, finally notice that because $\frac{1}{4} > \epsilon_{n} > 0$, $(1 - 2 \epsilon_{n}) \geq \frac{1}{2}$ for all $n = 1,2,3,...$
Clearly then, $\chi_n \in C_c(\mathbb{R}_{>1}) \subset C_0(\mathbb{R}_{>1})$ for each $n$, and we have $|| \sum_{k=m}^n \chi_{k} ||_{u} \leq \frac{1}{m}$, for all $1 \leq m < n$, so that $\sum_{k \geq 1} \chi_{k} \rightarrow_{||.||_{u}} f$, where $f \in C_0(\mathbb{R}_{>1})$ and is non-negative everywhere.
Writing $\mathbb{R}_{>1} = (1,2) \cup (2,3) \cup (3,4) \cup ...$, we see that $\sum_{k \geq 1} \chi_k \restriction (n,n+1) = \chi_n$, as $\text{supp}(\chi_{k}) \subset (k,k+1)$ for all $k \geq 1$, so that only $\chi_n$ may be non-zero when $f$ is restricted to $(n,n+1) \subset \mathbb{R}_{>1}$.
geetha290krm claims that the measures $\mu_n = n d \lambda \restriction (n,n+1) $, which are indeed in $\mathcal{M}(\mathbb{R}_{>1})$ converge to $0$ in the weak$^{\ast}$-sense.
This is false, as we will now show. Indeed, suppose it were true. Then by the definition of convergence in the weak$^{\ast}$-sense as defined by the OP, we should have $\int f d \mu_n \rightarrow_{n \rightarrow \infty} 0 = \int f d\mu$ , here $\mu = 0$ is the zero measure on $\mathbb{R}_{>1}$.
However, we have that $\int f d\mu_n \geq \int_{[n + \epsilon_n, n+1 - \epsilon_n]} f d\mu_n = (1 - 2 \epsilon_n)\frac{n}{n} \geq (1 - 2 \epsilon_n) \geq \frac{1}{2} $ for all $n = 1, 2 , 3 , ...$ with the inequalities all holding because $f$ is non-negative, its restriction to each interval $[n+\epsilon_n, n+1 - \epsilon_n] \subset (n,n+1)$ is just $\chi_n$ restricted to this interval, where it is identically equal to $\frac{1}{n}$, and because we chose $\frac{1}{4} > \epsilon_n > 0$ for all $n = 1,2,3,...$, we indeed have that $(1 - 2 \epsilon_n) \geq \frac{1}{2}$ for all $n = 1,2,3,...$ (as already established in the first paragraph).
This shows that we do not have $\int f d \mu_n \rightarrow_{n \rightarrow \infty} \int f d \mu$, so that geetha290krm's counterexample is invalid, because $\mu_n$ does not converge to $0$ in the weak$^{\ast}$ - sense, as I have demonstrated.
Indeed, there is no counterexample to (2), by the Riesz-Markov theorem , whenever $X$ is a locally compact Hausdorff space, since then your $\mathcal{M}(X)$ is precisely isometric to $C_0(X)^{\ast}$, i.e. the dual space to $C_0(X)$, consisting of all bounded $\mathbb{R}$-linear functionals $C_0(X) \rightarrow \mathbb{R}$, which is a Banach space with the operator norm. By equipping $C_0(X)^{\ast}$ with the weak$^{\ast}$-topology, and then transporting this topology onto $\mathcal{M}(X)$ via the isometric isomorphism $\mathcal{M}(X) \rightarrow C_0(X)$ given by $\mu \rightarrow I_{\mu}$, where $I_{\mu}(f) = (f \rightarrow \int_{X} f d \mu)$, $I_{\mu} : C_0(X) \rightarrow \mathbb{R}$, your question for (2) is answered in the affirmative, because it is a general fact that for a Banach space $X$, a weak$^{\ast}$-convergent sequence $f_n \in X^{\ast}$ is uniformly bounded in operator norm.
Therefore to seek a counterexample for (2), we need to find a (metrizable) Hausdorff topological space $X$, which is NOT locally compact.
One candidate is $X = (1,2) \times (2,3) \times (3,4) \times (4,5) \times ...$, with the usual product metric (inducing the product topology). This is NOT locally compact, so perhaps we could find a counterexample here.
Now perhaps it is possible to adapt geetha290krm's counterexample to this situation. But I have not figured out all the details yet and am not entirely convinced $X$ is sufficient to provide a counterexample to (2) for the following reason:
While this space $X$ is not locally compact, it is Borel isomorphic to $[0,1]$, which is indeed (locally) compact Hausdorff, so perhaps this $X$ may not work to produce a counterexample.
So for (2) a counterexample may need to come from a metrizable topological space $Y$, that is NOT locally compact, and is either
1: NOT second countable, or
2: NOT $\textbf{completely}$ metrizable if it $\textbf{is}$ second countable (i.e. if condition 1. does hold)
Either of these conditions will prohibit $Y$ from being Borel isomorphic to a Borel subset of $[0,1]$, where (in the latter space $[0,1]$) the usual Riesz-Markov theorem applies, and this may cause complications.
Best Answer
I suspect that when you wrote $|\mu_n|(\Omega)\to|\mu|(\Omega)$ maybe that's not exactly what you meant. Replying to the question as stated:
You assume so little about $\mu$ that it really has nothing to do with the rest of what's going on; there's no relationship between $\lambda$ and $\mu$. One doesn't need to "construct" a counterexample, any almost random choice of $\mu_n$ and $\mu$ works.
Say $\mu_n=\delta_{1/n}$, a point mass at $1/n$. So $\mu_n\to\delta_0$. Let $\Omega=(-1,1)$ and let $\mu$ be any positive measure with $\mu(\Omega)=1$.
Or maybe to better illustrate how $\mu$ really has nothing to do with $\lambda$, let $\Omega=(2,3)$ and let $\mu$ be any measure with $\mu(\Omega)=0$. Or $\Omega=\emptyset$ and $\mu=0$.
Heh, let $\mu_n$ be any norm-bounded sequence, $\Omega=\emptyset$, $\mu=0$.
Edit: Regarding the edit made to the question, and the comment asking whether there's any significant difference: Well of course there's a huge difference, since in the new version we have a specific "sequence" of measures! In particular, for example, if I'm reading things correctly $\mu_\epsilon$ is supported in the annulus $A_\epsilon=\{1-\epsilon\le|x|\le1\}$.
If I have the picture right it seems clear that the gradient of $u_\epsilon$ is $\nabla u_\epsilon(x)=-\frac1\epsilon\frac x{|x|}$ or something like that in $A_\epsilon$, $0$ elsewhere. So it seems clear that $\mu_\epsilon\to\lambda$, where $d\lambda=-n(x)\,dH$ (where $n$ is the outward unit normal on the sphere and $H$ is surface area on the sphere), which certainly appears to be ac wrt $H$.
Of course that could be all wrong, it's just my first impression. But note that the various things I say "seem clear" seem clear to me based on my picture of what $\mu_\epsilon$ actually is, not because of any general principle analogous to what you ask about in the original version of the question!
As a general rule, if they assert P and you don't see why P holds you might be better off actually stating what they actually assert and asking why it holds, instead of sort of guessing that they seem to be saying that P follows from Q and asking whether Q implies P.
Second edit: Two things.
(i) A conjecture regarding the sort of "soft" or "abstract" argument the authors might have had in mind: It's easy tp see that $||\mu_\epsilon||$ is bounded. And $\mu_\epsilon$ has a certain sort of rotational symmetry (which I'm not going to try to define precisely; this isn't my argument after all); hence any weak limit $\lambda$ must have the same symmetry. It's clear that $\lambda$ must be supported on $S=\{|x|=1\}$, since the support of $\mu_\epsilon$ shrinks to $S$, and the only vector-valued measure on $S$ with that symmetry is $cn\,dH$.
That's pretty vague; I'm not going to try to make it more precise, since it's just my guess regarding sort of what the authors may have had in mind. But:
(ii) Why it seems clear to me that $\mu_\epsilon$ simply does converge to what I say it does:
First, in general if $u$ is a radial function, $$u(x)=\phi(|x|),$$then $$\nabla u(x)=\phi'(|x|)\frac x{|x|}.$$("Advanced calculus": $\nabla u$ is the directional derivative in the direction of greatest increase, which is to say in a direction orthogonal to the level sets of $u$...) Hence $\nabla u_\epsilon$ is what I say it is above.
Now assume $f\in C_c(\Bbb R^n)$ and integrate in polar coordinates:
$$\int_{\Bbb R^n}f(x)\nabla u_\epsilon(x)\,dx=\int_S\frac{-1}{\epsilon}\int_{1-\epsilon}^1f(r\xi)\xi r^{n-1}\,drdH(\xi).$$And since $f$ is continuous it's clear that $$\lim_{\epsilon\to0}\frac{-1}{\epsilon}\int_{1-\epsilon}^1f(r\xi)\xi r^{n-1}\,dr=-f(\xi)\xi,$$uniformly over $\xi\in S$.
("Polar coordinates": In general $$\int_{\Bbb R^n}g(x)\,dx=c_n\int_S\int_0^\infty g(r\xi)r^{n-1}\,drdH(\xi).$$Note that if $H$ is actual "surface area" on $S$, in particular not normalized to be a probability measure as is sometimes done in that formula, then $c_n=1$. See Folland Real Analysis or various other places.)