Are vague convergence and weak convergence of measures both weak* convergence?

Ok so there is a lot to say here, let's start with the easiest question.

The use of $C^+_K$ instead of $C_K$ change absolutely nothing because we can always write $f=f^+-f^-$ with $f^+=\max(f,0)$ and $f^-=\max(-f,0)$ which are positive and integrable.

In order to avoid confusions between the different names of the different convergence i will talk about $C_K$, $C_0$ or $C_b$ convergence.

The simple case

First let's see what happen in the nicest case : Take $X$ to be a compact Hausdorff set. Then $C_K(X)=C_b(X)=C^0(X)$ with $C^0(X)$ the space of all continuous functions from $X$ to $\mathbb R$ (or $\mathbb C$, but it doesn't really matter). It's a bit difficult to say what $C_0(X)$ is for $X$ a topological space with no other structure but since here $X$ is compact the only natural definition is $C_0(X)=C_b(X)$. Moreover, using the Riesz-Markov theorem we can identify the (topological) dual of $C^0(X)$ and the space $\mathcal M (X)$ of all (signed and finite) Radon measures on $X$. So now we can define the weak-* topology on $\mathcal M (X)$ by $\mu_n\to \mu$ if and only if $\int fd\mu_n \to \int f\mu$ for every $f\in C^0(X)$. Obviously all the definitions you have seen before are the same here because $C_K(X)=C_0(X)=C_b(X)=C^0(X)$. So in this case all those different convergences are the same and are, in fact, the weak-* convergence of measures. The weak-* topology of a Banach has a very important propety : the closed unit ball for the strong topology is compact for the weak-* topology (it's the Banach alaoglu theorem). Which imply the following : given a bounded sequence of Radon measures $\mu_n$ there exists a Radon measure $\mu$ and a subsequence $\mu_{n_k}$ such that $\mu_{n_k}$ is weak-* convergent to $\mu$. Moreover we can prove that if the $\mu_n$ are probability measures then $\mu$ is also a probability measure (we'll see that later).

Now in the more general case when $X$ is Hausdorff but only localy compact it's more difficult. First of all the spaces $C_K$, $C_b$ and $C^0$ are all different. And when $C_0(X)$ is easily definable (when $X$ is a metric space or a topological vector space for example) it's also different from all the previous spaces. So the chances are that the different definitions of convergences won't coincide anymore. First let's see what are the differences with the previous case.

The things that still work, or not

If we look at the Riesz Markov theorem it still work with $X$ only being locally compact, but the Radon measure are not finite anymore, only locally finite. So the convergence using $C_K(X)$ functions is still the weak-* convergence of measures. The Banach Alaoglu theorem also still work but the property about probability measure doesn't hold anymore : in $\mathbb R$ the sequence $\delta_n$ is weak-* convergent to $O$, which is not a probability measure. However if the sequence $\mu_n$ is bounded and converge to $\mu$ then $\mu$ is also bounded and $\mu(X)\leq \lim\mu_n(X)$ (with the possibility of a strict inequality, as in the example before).

For the convergence with $C_0(X)$ functions we also have a representation theorem for the dual of $C_0(X)$ in terms of measures (see the wikipedia page of the Riesz-Markov theorem). So what i've said for $C_K(X)$ convergence will still hold. However even if the $C_K$ and the $C_0$ convergences can be viewed as weak-* convergences they are not the same. Take the sequence $n\delta_n$, it's $C_K$ convergent to $0$ but not $C_0$ convergent (take for example $X=\mathbb R$ and $f(x)=sin(x)/x$).

For the convergence with $C_b(X)$ functions it's different. There exists elements of the dual of $C_b(X)$ which cannot be represented as measures on $X$, for an exemple of that you can look at "Banach limit", the Banach limit of the $l^\infty$ sequence $(f(n))_{n\in\mathbb N}$ is one of those weird element of the dual of $C_b(X)$. So the $C_b$ convergence is not a weak-* convergence.

Last thing to say : since $C_K(X)\subset C_0(X) \subset C_b(X)$ we immediatly see that the $C_b$ convergence imply $C_0$ convergence which in turn imply $C_K$ convergence but the converses are all false. For example $\delta_n$ in $\mathbb R$ is $C_0$ convergent to $0$ but not $C_b$ convergent (take the bounded function sinus for example).

How to get things to work again

The almost obvious answer is : we have to work on a compact again. The usefull notion here is the notion of tight sequence (especially when dealing with probability): $(\mu_n)$ is called tight if, for all $\varepsilon>0$ there exists a compact $K_\varepsilon$ such that $\mu_n(K^C_\varepsilon)<\varepsilon$ for all $n$. So the sequence of measures are all almost supported in a compact set, so there is no possibility of mass "escaping at infinity" as it was the case with $(\delta_n)$. If $(\mu_n)$ is tight then the $\mu_n$ are finite (we assume that all the measures are locally finite)

Now you can read this answer of mine about tight sequences : Defining weak* convergence of measures using compactly supported continuous functions

To sum up things : if your bounded sequence of measures is tight then the 3 definitions are equivalent. Moreover if the $\mu_n$ are probability measure then so is the limit, and if you have a sequence of probability measure converging to a probability measure $\mu$ in the $C_K$ definition then it is tight. Moreover if $(\mu_n)$ is $C_b$ convergent then is is automatically tight.

The other questions

Locally finite measure that are note finite are usually not in the dual of $C_b$ and $C_0$. So the only definition of convergence that make sense here is the $C_K$ convergence. The definition 4 is consistent with the other definitions of vague convergence because the linear combinations of $\mathbf 1_{[a;b[}$ are dense in $C_0$ and because the sequence $(\mu_n)$ is bounded.

If you read french you can learn more on http://www.proba.jussieu.fr/cours/dea/telehtml/telehtml.html