I will concentrate on comparing (3) and (4). The definition (1) is meant for finite signed measures, whereas all the other definitions are meant for arbitrary positive measures; (1) is equivalent to (4) in the case of finite positive measures. (2) appears to be equivalent to (4) ["locally finite" can mean "finite on compact sets", although it is sometimes taken to mean "finite on the elements of some topological basis"; these are equivalent in the LCH (locally compact Hausdorff) case]. Finally, (5) does not appear to be a definition at all, but rather a description of a definition.
Now then,
i) In the case of a second countable LCH space, every locally finite measure satisfies both (3) and (4) (Theorem 7.8 of [1]). This is the most commonly considered scenario in applications, which is why almost no one bothers to carefully sort out the differences between the different definitions.
ii) In the case of a sigma-compact LCH space, (3) and (4) are equivalent. The forward direction is Corollary 7.6 of [1]; the backwards direction follows from the forward direction together with (iv) below (but I'm sure there is an easier proof).
iii) (3) and (4) are not equivalent in general, even for LCH metrizable spaces (Exercise 7.12 of [1]).
iv) In an LCH space, there is a bijection between
A) measures satisfying (3),
B) measures satisfying (4), and
C) positive linear functionals on the space of continuous functions with compact support.
(The Riesz representation theorem gives either (A)<->(C) or (B)<->(C), depending on where you look; (A)<->(B) is in the Schwarz book mentioned by Joe Lucke; see also Exercise 7.14 of [1])
[1] G. B. Folland, Real Analysis: Modern Techniques and Their Applications
Note: In [1], "Radon" refers to measures satisfying (3).
Edit: Below I assume that "measure" means "positive measure". Looking back at the question, it seems you are probably talking about real measures. The result for real measures follows, since any real measure is the difference of two finite positive measures.
If I'm not mistaken the answer to the question is yes (at least for finite $\mu$; recall that Radon measures are finite):
Suppose $\mu$ is a finite regular Borel measure on $[0,\omega_1]$. There exist $\alpha<\omega_1$, $a_j\ge 0$ for $0\le j\le\alpha$ and $c\ge0$ such that $\mu=\sum_{j\le\alpha}a_j\delta_j+c\delta_{\omega_1}$.
Proof: Let $E=\{j\in[0,\omega_1]:\mu(\{j\})>0\}$. Countable additivity shows that $E$ is countable. So there exists $\alpha<\omega_1$ with $$E\subset[0,\alpha]\cup\{\omega_1\}.$$Define $a_j=\mu(\{j\})$ for $j\le\alpha$ and $c=\mu(\{\omega_1\})$, and let $$\mu_1=\sum_{j\le\alpha}a_j\delta_j+c\delta_{\omega_1},$$ $$\nu=\mu-\mu_1.$$We need to show $\nu=0$. Note first that $\mu_1\le\mu$ (because $\mu_1(A)=\mu(A\cap E)$ [hint: $A\cap E$ is countable]), so $\nu\ge0$; hence it's enough to show that $\nu([0,\omega_1])=0$.
But $\nu(\{j\})=0$ for every $j\in[0,\omega_1]$, so countable additivity shows that$$\nu([0,\alpha])=0\quad(\alpha<\omega_1).$$If $K$ is a compact subset of $[0,\omega_1)$ there exists $\alpha<\omega_1$ with $K\subset[0,\alpha)$; hence $$\nu(K)=0\quad(\text{compact } K\subset[0,\omega_1)).$$So regularity shows that $\nu([0,\omega_1))=0$, and since $\nu(\{\omega_1\})=0$ we have $\nu([0,\omega_1])=0$.
Note. This is closely related to the result of Rao & Rao mentioned in the question. Indeed, we could have used Rao & Rao to obtain $\mu_1$ and $\nu$ above (I didn't do that because the construction is very simple a priori, and the version of Rao & Rao in the original version of the question was obviously false.) And in fact the result above suffices to reduce Rao & Rao to a special case:
Say $\mu$ is a (finite positive) Borel measure on $[0,\omega_1]$. Define $\mu_1$ and $\nu$ as above. Then $\mu_1$ is concentrated on a countable set, so we "only" have to show that $\nu$ is a multiple of the Dieudonne measure. So Rao & Rao is equivalent to this:
Lemma (Rao & Rao). If $\nu$ is a Borel probability measure on $[0,\omega_1]$ and $\nu(\{j\})=0$ for every $j\in[0,\omega_1]$ then $\nu$ is the Dieudonne measure.
I'm totally stuck on that, in fact it seems implausible. Edit: There's a proof here.
Best Answer
We shall call a space Lindelöf, if each its open cover has a countable subcover.
Proposition. Each $\sigma$-additive measure $\mu$ with discrete support $S$ defined on a $\sigma$-algebra (containing all Borel subsets of $X$) of subsets of a Lindelöf space $X$ is a sum of weighted Dirac measures centered at points of the support.
Proof. For each point $x\in X\setminus S$ pick a neighborhood $U_x$ with zero measure. An open cover $\mathcal U=\{U_x: x\in X\setminus S\}$ has a countable subcover $\mathcal V$. Then $$\mu(X\setminus S)=\mu\left(\bigcup\mathcal U\right)=\mu\left(\bigcup\mathcal V\right)=\sum_{V\in\mathcal V}\mu(V)=0.$$ For each point $x\in S$ pick a neighborhood $V_x$ such that $V_x\cap S=\{x\}$. Then $$\mu(V_x)=\mu(V_x\cap S)+\mu(V_x\cap (X\setminus S))=\mu(\{x\}).$$ Since $\{V_x\cap S: x\in S\}$ is a disjoint open cover of a Lindelöf space $S$, it is countable. Now let $A$ be any $\mu$-measurable subset of $X$. Then $$\mu(A)=\mu(A\cap S)+\mu(A\cap (X\setminus S))=\sum_{x\in A\cap S}\mu(\{x\}).$$ $\square$