*(In order that this question appears as answered I'm converting Darij Grinberg's comment into an answer.)*

For $j$ and $k$ (that will remain fixed) in $\mathbb N$, and a set $X$, denote $\mathcal F(X):=\mathcal{P}_k\mathcal{P}_j(X)$, the set of all $k$-element sets $\mathcal U:=\{U_1,\dots,U_k\}$ of $j$-element subsets of $X$. Denote $\mathcal C(X)\subset \mathcal F(X)$ the set of those elements $\mathcal U$ of $\mathcal F(X)$ which are coverings of $X$, that is $\displaystyle \bigcup_{U\in \mathcal U}U =X$. Every $\mathcal U\in\mathcal F(X)$ is a covering of exactly one subset $Y$ of $X$, namely $Y:=\displaystyle \bigcup_{U\in \mathcal U}U$. Therefore
$$\mathcal F(X)=\bigsqcup_{Y\subset X } \mathcal C(Y)$$
$$\mathcal C(X)=\Big(\bigcup_{x\in X } \mathcal F\big(X\setminus\{x\}\big)\Big)^c,$$
and note also that
$$ \bigcap_{x\in Y} \mathcal F\big(X\setminus\{x\}\big)= \mathcal F\big(X\setminus Y\big).$$
As to cardinalities, if $|X|=n$, we have of course $\big|\mathcal F(X)\big|=\begin{pmatrix} n\\j \\ k \end{pmatrix} := \bigg( {{n\choose j}\atop k}\bigg) $, and, if we denote $\begin{bmatrix} n\\j \\ k \end{bmatrix}:=\big|\mathcal C(X)\big| $ we have from the above disjoint union
$$\begin{pmatrix} n\\j \\ k \end{pmatrix}=\sum_{m=0}^n{n\choose m}\begin{bmatrix} m\\j \\ k \end{bmatrix},$$
which can be inverted giving
$$\begin{bmatrix} m\\j \\ k \end{bmatrix}=\sum_{n=0}^m(-1)^{m-n}{m\choose n}\begin{pmatrix} n\\j \\ k \end{pmatrix}.$$
The latter, of course, can be seen as an instance of the inclusion-exclusion formula: $$\big|\mathcal C(X)\big| =\bigg|\Big(\bigcup_{x\in X } \mathcal F\big(X\setminus\{x\}\big)\Big)^c\bigg|=\sum_{Y\subset X}(-1)^{|Y|}\big|\mathcal F\big(X\setminus Y\big)\big|=\sum_{Y\subset X}(-1)^{|X\setminus Y|}\big|\mathcal F(Y)\big|$$