Martin has already explained the notation, but you might also find the following connection with Stirling numbers of the second kind useful, since those are the ones mentioned in your title.
If you calculate the $c(n,k)$ for $n=0,1,2,3,4,5$, you get the following array of $0$-th diagonals:
n\k: 0 1 2 3 4 5
---------------------------
0 | 1
1 | 0 1
2 | 0 1 2
3 | 0 1 6 6
4 | 0 1 14 36 24
5 | 0 1 30 150 240 120
This triangle can be found as A131689 at OEIS, where you can discover that the entry in row $n$, column $k$ is $$k!\left\{\matrix{n\\k}\right\},$$ where $\displaystyle\left\{\matrix{n\\k}\right\}$ is the Stirling number of the second kind. And sure enough, if we divide each entry in column $k$ by $k!$ we get the following table:
n\k: 0 1 2 3 4 5
---------------------------
0 | 1
1 | 0 1
2 | 0 1 1
3 | 0 1 3 1
4 | 0 1 7 6 1
5 | 0 1 15 25 10 1
This is readily recognizable as the table of Stirling numbers of the second kind.
explain this "identical boxes"
When we put distinguishable objects into "identical boxes", it means we put distinguishable objects into distinguishable boxes, but then we want to consider rearranging the boxes to arrive at an equivalent arrangement, and we count the number of equivalence classes.
A better way of thinking about it is that we have a set of boxes (not a list), and each box is a nonempty set. So the order of the boxes doesn't matter, and the order of objects within each box doesn't matter.
Perhaps an example will help.
The number of ways to put $n$ distinguishable objects $1,2,3,\ldots, n$ into $k$ identical boxes when $n = 4$ and $k = 2$ is $7$:
Box 1 Box 2
123 4
124 3
134 2
234 1
124 3
12 34
13 24
14 23
1 how to construct the formula for it (the recursive one) preferably using bijection principle and just using "box and objects language"
Let's prove the recurrence relation
$$
\newcommand{\stir}{\genfrac\{\}{0pt}{}}
\stir{n+1}{k} = k \stir{n}{k} + \stir{n}{k-1}.
$$
Consider an arrangement of $n+1$ objects $(1,2,3,\ldots,n+1)$ put in $k$ boxes, of the kind counted by $\stir{n+1}{k}$.
There are two cases to consider: either object $n+1$ stands in its own box, or it shares its box with some other objects.
If it stands in its own box, then we have to distribute the remaining $n$ objects into the remaining $k-1$ boxes, and there are $\stir{n}{k-1}$ ways to do so. If not, then we have to put $n$ remaining objects into $k$ boxes, one of them distinguished from the rest, such that each box is nonempty (the box that contained object $n+1$ has to be nonempty too, as that is the case we are considering). The number of ways to do this is $k \cdot \stir{n}{k}$: first $\stir{n}{k}$ ways to put the $n$ objects into $k$ identical boxes, and then $n$ ways to choose which box to select as the distinguished one, which we will put $n+1$ into. (Note that nonempty is key here: because all boxes are nonempty, despite being indistuishable boxes they are distinguishable once any object is put in them; thus, we cannot put object $n+1$ into two different boxes and end up with the same resulting arrangement.)
2 explain the same but now with " set partition language"
In a partition of $1,2,3,\ldots,n+1$ into $k$ sets, there are two possibilities: either $\{k\}$ is an element of the partition, or $\{k\}$ is not an element of the partition. In the forms case,
there is a one-to-one, onto map (bijection) from all such partitions of $1,2,3,\ldots,n+1$ into $k$ parts to all partitions of $1,2,3,\ldots,n$ into $k-1$ parts, given by
(for $P$ a partition of $\{1,2,3,\ldots,n+1\}$
$$
P \quad\mapsto\quad P \setminus \{\{n+1\}\}.
$$
In the latter case, there is a $k$-to-one, onto map from all such partitions of $1,2,3\ldots,n+1$ into $k$ parts to all partitions of $1,2,3, \ldots, n$ into $k$ parts, given by
$$
P \quad\mapsto\quad \{x \setminus \{n+1\} \;|\; x \in P\}.
$$
3 define and construct the explicit formula for it
Explicit formulas are not always that nice :)
You can find a combinatorial answer to building an explicit formula here.
Best Answer
The book Concrete Mathematics by Graham, Knuth & Patashnik suggests “$n$ subset $k$” for ${n\brace k}$ and “$n$ cycle $k$” for ${n \brack k}$ (pp. 258–259 in the second edition).