I just have some questions in regards to how equations of sets work.

The operation $X\oplus Y$ is called the symmetric difference of sets $X$ and $Y$ (often denoted instead by $X\triangle Y$), and the reason you can always do the given replacement is because it is defined as $$X\oplus Y\ :=\ (X\setminus Y)\cup (Y\setminus X)$$ which is the same as $(X\cap Y^c)\cup(Y\cap X^c)$ by the definition of set difference $\setminus$.

An element $x$ is in $X\oplus Y$ iff either $x\in X$ or $x\in Y$ but not both.

It seems that you're missing the weapon of the distributivity property, which states that both $\cap$ and $\cup$ behave to each other like multiplication to addition: $x\cdot(y+z)=x\cdot y+x\cdot z$: $$X\cap(Y\cup Z)=(X\cap Y)\,\cup\,(X\cap Z)\\ X\cup(Y\cap Z)=(X\cup Y)\,\cap\,(X\cup Z) $$ Draw the Venn diagram for these equalities and also prove them by picking an element from each side and showing that it's also in the other side.

Actually, already for the symmetric difference we have $$X\oplus Y\ =\ (X\cap Y^c)\,\cup\,(Y\cap X^c)\ =\ \big((X\cap Y^c)\cup Y\big)\,\cap\,\big((X\cap Y^c)\cup X^c\big)\ =\\ =\ \big((X\cup Y)\,\cap\, (Y^c\cup Y)\big)\,\cap\,\big((X\cup X^c)\cap (Y^c\cup X^c)\big)\ = \\ =\ (X\cup Y)\,\cap\,(Y^c\cup X^c)\ =\ (X\cup Y)\,\cap\,(X\cap Y)^c\ =\ \boxed{ (X\cup Y)\setminus(X\cap Y)} $$ which result is straightforward to see from the Venn diagram or from the characterization by elements.

Important properties of the symmetric difference:

• $X\oplus X=\emptyset$
• $X\oplus\emptyset=X$
• $X\oplus(Y\oplus Z)=\{$elements that are present in an odd number of the three sets $X,Y,Z\}=(X\oplus Y)\oplus Z$
• intersection distributes over it: $X\cap(Y\oplus Z)=(X\cap Y)\oplus (X\cap Z)$.
• $X^c\oplus Y^c = X\oplus Y$ (Try to prove these.)

As a consequence, if we have an equation of sets $L=R$, then it is indeed equivalent to $L\oplus R=\emptyset$, because $$L=R\,\implies\, L\oplus R=R\oplus R=\emptyset\, \implies\, \\ L=L\oplus\emptyset=L\oplus(R\oplus R)=(L\oplus R)\oplus R=\emptyset\oplus R=R\,.$$

Now, to the equations. I think, by using distributivity and properties like $X\cap X^c=\emptyset$ you can finish up your work.
However, we can apply simplifications to make life easier, especially the above mentioned distributivity can be handy, e.g. here we 'extract $B$' just like how $b$ is extracted from the sum of numbers $bx'+ba'=b(x'+a')$: $$(B\cap X^c)\oplus(B\cap A^c)=B\cap(X^c\oplus A^c)=B\cap(X\oplus A)=\\ =B\cap\big((X\setminus A)\cup (A\setminus X)\big)=\ (B\cap X\cap A^c)\,\cup\, (B\cap A\cap X^c)$$ written as a union of intersections.

So the second equation holds iff $$X\,\cap\,B\cap A^c=\emptyset\text{ and }X^c\,\cap\,A\cap B=\emptyset$$ that is, iff $$X\subseteq (B\cap A^c)^c=B^c\cup A\text{ and }X^c\subseteq(A\cap B)^c$$ iff $$A\cap B\ \subseteq \ X\ \subseteq\ A\cup B^c\,.$$