Consider a finitely complete category $\mathscr{B}$. A universal closure operation on $\mathscr{B}$ consists in giving, for every subobject $S\rightarrowtail B$ in $\mathscr{B}$, another subobject $\overline{S}\rightarrowtail B$ called the closure of $S$ in $B$; these assignments have to satisfy the following properties, where $S,T$ are subobjects of $B$ and $f:A\longrightarrow B $ is a morphism of $\mathscr{B}$:
1) $S\subseteq \overline{S}$
2) $S\subseteq T\Rightarrow \overline{S}\subseteq\overline{T}$
3) $\overline{\overline{S}}=\overline{S}$
4) $f^{-1}(\overline{S})=\overline{f^{-1}(S)}$
$\textbf{Questions:}$
1) how $\subseteq$ is defined? My guess is that a subobject $s:S\longrightarrow B$ is $\subseteq $ than another subobject $t:T\longrightarrow B$ when there exists a morphism $f:S\longrightarrow T$ such that $t\circ f=s$. Is it correct?
2) What is $f^{-1}$? I mean, the notation is the same as preimage, but what does it mean preimage in this contest?
For (1), you are right.
For (2): if $S \to B$ is a subobject, and $f: A \to B$ is any morphism, then we can form (because $\mathscr B$ is finitely complete) the pullback $$ \require{AMScd} \begin{CD} f^{-1}(S) @>>> S\\ @VVV @VVV \\ A @>{f}>> B. \end{CD} $$ Because monomorphisms are stable under pullback, $f^{-1}(S) \to A$ is a subobject. You can check for yourself that in e.g. $\mathsf{Set}$, this corresponds to the usual notion of inverse image of a subset.