The connection between your example and the more general definition is that $\bigcup\{a,b\}=a\cup b$. Written out in all its gory details, this is
$$\bigcup\Big\{\{1,2,3\},\{4,5\}\Big\}=\{1,2,3\}\cup\{4,5\}=\{1,2,3,4,5\}\;.$$
Let’s check that against the definition:
$$\begin{align*}
&\bigcup\Big\{\{1,2,3\},\{4,5\}\Big\}\\
&\qquad=\left\{x:\text{there exists an element }y\in\Big\{\{1,2,3\},\{4,5\}\Big\}\text{ such that }x\in y\right\}\\
&\qquad=\Big\{x:x\in\{1,2,3\}\text{ or }x\in\{4,5\}\Big\}\\
&\qquad=\{1,2,3\}\cup\{4,5\}\\
&\qquad=\{1,2,3,4,5\}\;.
\end{align*}$$
Take a slightly bigger example. Let $a,b$, and $c$ be any sets; then
$$\begin{align*}
\bigcup\{a,b,c\}&=\Big\{x:\text{there exists an element }y\in\{a,b,c\}\text{ such that }x\in y\Big\}\\
&=\{x:x\in a\text{ or }x\in b\text{ or }x\in c\}\\
&=a\cup b\cup c\;.
\end{align*}$$
One more, even bigger: for $n\in\Bbb N$ let $A_n$ be a set, and let $\mathscr{A}=\{A_n:n\in\Bbb N\}$. Then
$$\begin{align*}
\bigcup\mathscr{A}&=\Big\{x:\text{there exists an }n\in\Bbb N\text{ such that }x\in A_n\Big\}\\
&=\{x:x\in A_0\text{ or }x\in A_1\text{ or }x\in A_2\text{ or }\dots\}\\
&=A_0\cup A_1\cup A_2\cup\dots\\
&=\bigcup_{n\in\Bbb N}A_n\;.
\end{align*}$$
In the usual set-theoretic construction of $\mathbb{N}$, in which $0 = \{\}$ and $n+1 = n \cup \{n\}$, you have
$$
\mathbb{N}=\{0,1,2,3,\ldots\}=\{\{\},\{0\},\{0,1\},\{0,1,2\},\ldots\},
$$
and so
$$
\bigcup\mathbb{N}=\{0,1,2,3,\ldots\}=\mathbb{N}.
$$
There's not a comparably simple construction of $\mathbb{R}$ in ZFC, so I think $\bigcup\mathbb{R}$, while it must exist, will depend heavily on the details. For instance, there are at least two different constructions in which the elements of $\mathbb{R}$ are equivalence classes (subsets) of another set: the set of Cauchy sequences in one case, and the set of Dedekind cuts in the other. Using these constructions, $\bigcup\mathbb{R}$ is either the set of the Cauchy sequences or the set of Dedekind cuts, which are clearly quite different.
Indeed, even $\bigcup\mathbb{N}=\mathbb{N}$, while elegant, isn't necessary. Other constructions of $\mathbb{N}$ include representing its elements as collections of finite subsets of some other infinite set $A$. (I.e., "3" is the set of all 3-element subsets of $A$, "4" is the set of all 4-element subsets, etc.) In this case, $\bigcup\mathbb{N}$ would be the set of all finite subsets of $A$.
Best Answer
An example may help. Let $$ F = \{(1, 2), (2, 3), (3, 1)\} = \{\{\{1\}, \{1, 2\}\}, \{\{2\}, \{2, 3\}\}, \{\{3\}, \{3, 1\}\}\} $$ Then: $$ \begin{align*} \bigcup F &=\{\{1\}, \{1, 2\}, \{2\}, \{2, 3\}, \{3\}, \{3, 1\}\} \end{align*} $$ And
$$ \begin{align*} \bigcup\bigcup F &= \bigcup\{\{1\}, \{1, 2\}, \{2\}, \{2, 3\}, \{3\}, \{3, 1\}\} \\ &= \{1, 1, 2, 2, 2, 3, 3, 3, 1\} \\ &= \{1, 2, 3\} \end{align*} $$