We will only prove here that definition (1) implies definition (2) (this is the easy part).
Suppose that the condition in definition (1) holds. That is, $\mu_n ((a,b]) \to \mu ((a,b])$ for any $a,b \in D$, $a < b$, where $D$ is a dense subset of $R$. Define $\tilde D = \lbrace x \in R:\mu (\lbrace x \rbrace ) = 0 \rbrace$. Then $\tilde D$ is dense in $R$, since its complement is at most countable. For any $a,b \in \tilde D$, $a < b$, and $a',a'',b',b'' \in D$, $a' < b'$, $a'' < b''$, such that $a'' < a < a'$ and $b' < b < b''$, we have
$$
\mu _n ((a',b']) \le \mu _n ((a,b)) \le \mu _n ((a'',b'']).
$$
By assumption, $\mu _n ((a',b']) \to \mu ((a',b'])$ and $\mu _n ((a'',b''] \to \mu ((a'',b''])$ as $n \to \infty$. Then, since $a',a''$ and $b',b''$ can be chosen arbitrarily close to $a$ and $b$, respectively, we can conclude that
$$
\mu ((a,b)) \le \mathop {\lim \inf }\limits_{n \to \infty } \mu _n ((a,b)) \le \mathop {\lim \sup }\limits_{n \to \infty } \mu _n ((a,b)) \le \mu ([a,b]).
$$
However, by assumption $\mu (\lbrace a \rbrace) = \mu (\lbrace b \rbrace) = 0$, and hence $\lim _{n \to \infty } \mu _n ((a,b)) = \mu ((a,b))$. Thus, the condition in definition (2) is satisfied (with $\tilde D$ playing the role of $D$ in that definition).
It is instructive to consider here the following simple example. As is customary, denote by $\delta_x$ the probability measure concentrated at $x$. Define $\mu_n = \delta_{-1/n}$, $n \geq 1$, and $\mu = \delta_0$. Then, for any $a,b \in R$, $a < b$, we have $\mu _n ((a,b]) \to \mu ((a,b])$, and hence, by definition (1), $\mu_n$ converges vaguely to $\mu$. On the other hand, $\mu _n ((-2,0)) = 1$ for any $n \geq 1$, whereas $\mu ((-2,0)) = 0$. However, if we define $\tilde D$ as above, then $\tilde D = R - \lbrace 0 \rbrace$, and for any $a,b \in \tilde D$, we do have $\mu _n ((a,b)) \to \mu ((a,b))$; hence, $\mu_n$ converges vaguely to $\mu$ also according to definition (2).
Finally, as for the OP's remark (concerning definition (2)) that "the original text just says converges, not mention that converges to $ \mu((a,b))$ which I guess can be added?", it is clear that we are deliberately only given that $\mu_n ((a,b))$ converges -- this is what makes the converse implication (from definition (2) to (1)) a substantially more difficult problem.