$\newcommand{\ran}{\operatorname{ran}}$Here’s one way to think about it. Suppose that $y\in\ran f$; then we can pull $y$ back to $f^{-1}(y)\in X$. If $f^{-1}(y)\in\ran g$, we can pull it back to $g^{-1}(f^{-1}(y))\in Y$. If we continue this pulling back, one of two things must happen: either we reach a dead end at a point of $X$ or $Y$ that can’t be pulled back (because it’s in $Y\setminus\ran f$ or $X\setminus\ran g$), or we don’t.
Let $X_0=X\setminus\ran g$, the set of points of $X$ that cannot be pulled back at all, and let $Y_0=Y\setminus\ran f$. More generally, for each $n\in\omega$ let $X_n$ be the set of points of $X$ that can be pulled back exactly $n$ times, and let $Y_n$ be the set of points of $Y$ that can be pulled back exactly $n$ times. Finally, let $X_\omega$ and $Y_\omega$ by the subsets of $X$ and $Y$, respectively whose points can be pulled back infinitely many times.
At this point a sketch helps; it should show the partitions $\{X_n:n\le\omega\}$ of $X$ and $\{Y_n:n\le\omega\}$ of $Y$, and it should include arrows indicating what parts of $X$ get mapped to what parts of $Y$ and vice versa. To avoid having arrows crossing, I’ve taken $X$ and $Y$ apart in the following diagram.
$$\begin{array}{}
X_0&\overset{f}\longrightarrow&Y_1&\overset{g}\longrightarrow&X_2&\overset{f}\longrightarrow& Y_3&\overset{g}\longrightarrow&X_4&\dots&X_\omega\\
Y_0&\overset{g}\longrightarrow&X_1&\overset{f}\longrightarrow&Y_2&\overset{g}\longrightarrow&X_3&\overset{f}\longrightarrow&Y_4&\dots&Y_\omega
\end{array}$$
Each of the arrows is a bijection, so I can break up the diagram into $\omega$ self-contained parts. The first two parts are:
$$\begin{array}{}
X_0&\overset{f}\longrightarrow&Y_1\\
Y_0&\overset{g}\longrightarrow&X_1
\end{array}\qquad
\begin{array}{}
X_2&\overset{f}\longrightarrow&Y_3\\
Y_2&\overset{g}\longrightarrow&X_3
\end{array}$$
Ignoring $X_\omega$ and $Y_\omega$ for the moment, I can rearrange the rest of the diagram to give my a bijection from $X\setminus X_\omega$ to $Y\setminus Y_\omega$:
$$\begin{array}{ccc}
X_0&\overset{f}\longrightarrow&Y_1\\
X_1&\overset{g^{-1}}\longrightarrow&Y_0\\
X_2&\overset{f}\longrightarrow&Y_3\\
X_3&\overset{g^{-1}}\longrightarrow&Y_2\\
\vdots&\vdots&\vdots\\
X_{2k}&\overset{f}\longrightarrow&Y_{2k+1}\\
X_{2k+1}&\overset{g^{-1}}\longrightarrow&Y_{2k}\\
\vdots&\vdots&\vdots
\end{array}$$
Finally, I claim that $f[X_\omega]=Y_\omega$: everything in $X_\omega$ can be pulled back infinitely often, so everything in $f[X_\omega]$ can be pulled back infinitely often, and therefore $f[X_\omega]\subseteq Y_\omega$. On the other hand, if $y\in Y_\omega$, then $y$ can be pulled back infinitely often, so it must be possible to pull $f^{-1}(y)$ back infinitely often, and therefore $f^{-1}(y)\in X_\omega$. Thus, $Y_\omega\subseteq f[X_\omega]$ as well. The diagram above can now be completed to show a bijection from $X$ onto $Y$:
$$\begin{array}{ccc}
X_0&\overset{f}\longrightarrow&Y_1\\
X_1&\overset{g^{-1}}\longrightarrow&Y_0\\
X_2&\overset{f}\longrightarrow&Y_3\\
X_3&\overset{g^{-1}}\longrightarrow&Y_2\\
\vdots&\vdots&\vdots\\
X_{2k}&\overset{f}\longrightarrow&Y_{2k+1}\\
X_{2k+1}&\overset{g^{-1}}\longrightarrow&Y_{2k}\\
\vdots&\vdots&\vdots\\
X_\omega&\overset{f}\longrightarrow&Y_\omega
\end{array}$$
The bijection is defined piecewise, but that’s no problem.
There are a few details to be filled in to make this a fully rigorous proof, but I think that it does give a reasonable idea of one possible intuition.
Added: Here’s a very rough sketch. Arrows from left to right are (parts of) $f$, and arrows from right to left are (parts of) $g$.
Best Answer
First lets check, that $g$ indeed maps to from $B' \to A'$. Suppose $g(b') \in A_n=g(f(A_{n-1}))$ for some $n$. So by injectivity we would have $b' \in f(A_{n-1})$, but this is already a contradiction.
Now why is $g$ restricted to $B'$ onto? Suppose there is an element $a' \in A'$ that is not hit by any $b' \in B'$. Since $a'$ cannot be in $A_1$ there has to be an element $b \in f(A_n)\subset B$ s.t. $g(b)=a'$. Since $b \in f(A_n)$ we can write it as $f(a)=b$ and therefore $a'=g(f(a))\in A_{n+1}$. But this is a contradiction to where $a'$ lives.
I'm also not sure why the disjointness is of importance. You could argue that the $f(A_n)$ in which $b$ lies is only unique if this is the case. But this doesn't seem to matter much for the argument to work.