Let $\alpha$ be an admissible ordinal. A set is $\alpha$-finite iff it is an element of $L_\alpha$. $A\leq_\alpha B$ iff there are $R_1$ and $R_2$ which are $\Sigma_1$-definable over $L_\alpha$ using parameters in $L_\alpha$ such that for all $\alpha$-finite sets $K$, $K\subseteq A$ iff there exists $\langle P_1,N_1,K\rangle\in R_1$ with $P_1\subseteq B,$ $N_1\subseteq \overline{B}$ and $K\subseteq \overline{A}$ iff there exists $\langle P_2,N_2,K\rangle\in R_2$ with $P_2\subseteq B,$ $N_2\subseteq \overline{B}.$ $R_1$ and $R_2$ are called $\alpha$-reductions.
When $\alpha<\beta$ are admissible, it could happen that $A\leq_\alpha B$ and $A\not\leq_\beta B$ for a trivial reason: $B=\emptyset,$ $A\cap L_\alpha=\emptyset$ (so $A\leq_\alpha B$) and $A\cap L_\beta\setminus L_\alpha$ is complicated (so $A\not\leq_\beta B$). This is not very illuminating, since $L_\alpha$ cannot even refer to any of the elements of $A$.
There is a more interesting example in which $A$ and $B$ are subsets of $\omega$. Let $\alpha$ be $\omega_1^{CK}$ and $\beta$ be the first admissible greater than $\omega_1^{CK}$. Fix a recursive pairing function according to which any $S\subseteq\omega$ can be viewed as a recursive join $S=\bigcup_{i\in\omega}(S)_i$. Let $G$ be an element of $L_\beta$ such that $G$ is Cohen generic over $L_\alpha$. Next, let $H_1$ and $H_2$ be mutually Cohen generic over $L_\beta$. Define $A=\bigcup_{i\in\omega}(A)_i$, where $(A)_i=(H_1)_i$ if $i\in H_2$ and $(A)_i=(G)_i$ if $i\not\in H_2$. Let $B$ be generic over $L_\beta[H_1,H_2]$ so that for all $n$, $n\in A$ iff either both of $2n$ and $2n+1$ belong to $B$ or neither of them does. The forcing partial order for $B$ is by finite conditions that satisfy the coding constraint. In other words, $A$ is a join of sets that are generic over $L_\alpha$, which are determined in a way that is generic over $L_\beta$ to be either components of $G$ or generic over $L_\beta$. $B$ is a yet more generic coding of the characteristic function of $A$ by a elementary truth table reduction. There are three things to check.
First, for any $K$ in $L_\alpha$, if $K\subseteq A$ or $K\subseteq \overline{A}$ then $K\cap\omega$ is finite. This is a typical feature of Cohen forcing--no finite condition can force the generic to satisfy an infinite inclusion. It follows that $A\leq_\alpha B$.
Second, for any $K$ in $L_\beta$, if $K\subseteq B$ or $K\subseteq \overline{B}$ then $K\cap \omega$ is finite. Here it is enough to observe that $B$'s coding an atomic statement about $A$ does not require that any particular number belong to $B$ or that it belong to $\overline{B}$. Then, apply genericity to avoid containing any infinite subset from $L_\beta$ in either $B$ or $\overline{B}$.
Finally, $A\not\leq_\beta B$. Here is a sketch of the proof for this last point. Consider $\beta$-reductions $R_1$ and $R_2$ and a condition $p_0$. Fix a condition $p$ stronger than $p_0$ on $H_1,H_2$ which specifies that $(A)_i=(G)_i$, where $p_0$ did not specify whether this holds (did not specify $H_2$ at $i$). If there do not exist $\langle P,N,(G)_i\rangle$ in $R_1$ and a condition $q$ extending $p$ such that $q$ forces $P\subseteq B$ and $N\subseteq\overline{B}$, then $p$ forces that $R_1$ and $R_2$ do not show that $A\leq_\beta B$. Otherwise, fix such $P,N$ and $q$. Both $P$ and $N$ must have finite intersection with $\omega$. Now, change $q$ to $q^*$ so that $q^*$ specifies that $(A)_i$ is generic over $L_\beta$ (that is, change what $q$ says about $H_2$ at $i$) and so that $q^*$ specifies that $(H_1)_i$ is equal to $(G)_i$ at every value that is mentioned in $P$ or in $N$. Then, $q^*$ is stronger than $p_0,$ forces that $(G)_i$ is not a subset of $A$ and so forces that $R_1$ and $R_2$ do not show that $A\leq_\beta B$.
