If $f$ is a bounded, measurable real function, and $\mu_n\to\mu$ in total variation, then $\int f \, d\mu_n\to \int f\, d\mu$. The reason is just that $f$ can be uniformly approximated by simple functions.
If $f$ is not bounded $\int f\, d\mu_n$ need not converge to $\int f\, d\mu$. For a counterexample, look at measures on the real line, let $\mu_n$ give $\{n\}$ probability $\frac1n$ and $\{0\}$ probability $1-\frac1n$, let $\mu$ give $\{0\}$ probability $1$, and let $f$ be the identity. Then $\mu_n\to\mu$ in total variation but, for all $n$, $\int f\, d\mu_n=1\ne \int f \,d\mu=0$.
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.)
Best Answer
I'm going to try to mimick the proof of Corollary 4.2 in the paper linked by @Clement C. Comments and generalizations are appreciated.
We first suppose that $\{\mu_n \}$ is a sequence of probability measures on $(X, \mathcal{F})$ that converges setwise to the probability measure $\mu$. Moreover, suppose $\mu_n(A) = \int f_n d \lambda$ and $\mu(A) = \int_A f d \lambda$, where $\lambda$ is a $\sigma$-finite measure on $(X, \mathcal{F})$ and $f_n \to f$ a.e. ($\lambda$).
Now, setwise convergence implies $\int f_n d\lambda \to \int f d\lambda$. By Scheffe's Lemma, $\int |f_n - f|d\lambda \to 0$. And this in turn implies $\|\mu_n - \mu \| \to 0$.