$L\_{\omega\_1}$ is the $\omega\_1$-th constructible hierarchy.
I define two binary relation on $P(\omega\_1)$ as follows:


For $X,Y\in{P}(\omega\_1)$,


$R\_1(X,Y)$ means: there is $U\subset\omega\_1\times\omega\_1$ which is $\Sigma\_1$ definable with parameters in $L\_{\omega\_1}$ such that $Y=\lbrace x\in\omega\_1\mid\exists{t}\in{X}(t,x)\in{U}\rbrace$.


$R\_2(X,Y)$ means: $Y$ is $\Sigma\_1(X)$ in the model $(L\_{\omega\_1},\in)$, i.e. $Y$ is $\Sigma\_1$-definable with $X$ as a unary predicate.


The question is:


1. $A$ is a subset of $\omega\_1$, define $A'$ as follows:
$A'$ is the set of all pairs $(a,b)$, $a,b\in\omega\_1$, $K(a)\subset{A}$ and $K(b)\subset\omega\_1\setminus{A}$.


($K:\omega\_1\rightarrow{L\_{\omega\_1}}$ is the canonical emumeration of elements of $L\_{\omega\_1}$.)


Then is $R\_1(A,A')$ true?
2. Are $R\_1$ and $R\_2$ equivalent?
3. Is $R\_1$ or $R\_2$ transitive?



About Sigma_1 definability
If I've followed your definitions, then the answer to question 1 is negative. 


Specifically, if $A$ is $\Sigma\_1$ but not $\Pi\_1$, then $R\_1(A,A')$ does not hold. The reason is that when $R\_1(A,A')$ holds, we get $\Sigma\_1$ information about the complement of $A$. To see what I mean, suppose $R\_1(A,A')$ holds, via the $\Sigma\_1$ definable set $U$. This means that $(a,b)\in A'$ if and only if $\exists t\in A\ (t,(a,b))\in U$. Furthermore, $\gamma\notin A$ if and only if $\exists (a,b)\in A'\ \gamma\in K(b)$, and this holds if and only if $\exists t\in A\ \exists (a,b)\ (t,(a,b))\in U$ and $\gamma\in K(b)$, which is a $\Sigma\_1$ definition of the complement of $A$, contrary to the assumption that $A$ was not $\Pi\_1$. 


The same idea shows that $R\_1$ and $R\_2$ are not equivalent, since $R\_2(A,A')$ does hold, as you can define $A'$ in a $\Sigma\_1$ way using $A$ as an atomic predicate: $(a,b)\in A'$ if and only if every element of $K(a)$ is in $A$ and every element of $K(b)$ is not in $A$. This is $\Sigma\_1(A)$, since we need one existential quantifier to get access to the values of $K(a)$ and $K(b)$, and then we can use bounded quantifiers to refer to the elements of these sets.


$R\_2$ is not transitive, since by applying it several times allows us to successively take complements and projections, thereby allowing us to reach any $\Sigma\_n$ definable set in $n$ steps, rather than merely the $\Sigma\_1$ definable sets. 


