Let $$A = \begin{bmatrix} 1 & 1 \\ 1 & 0\\ \end{bmatrix}.$$ Then the eigenvalues of $A$ are $1/2(1+\sqrt{5})$ and $1/2(1-\sqrt{5})$. The eigenvector corresponding to the unstable eigenvalue is the line $y = 1/2(-1+\sqrt 5)x$, whereas the stable eigenvalue has eigenvector $y = -1/2 (1+\sqrt 5)x$.
Let $L$ be a hyperbolic linear automorphism of the torus induced by $A$.
Could someone explain how we get the following graph? To simplify the question, let's jsut discuss how $L$ affects $R_1$. The upper left picture is a Markov partition of the torus. The dark regions on the second and third pictures represent $R_1$, $L(R_1)$ and $L^{-1}(R_1)$.
Since the edges of your rectangles are exactly the eigendirections, $L$ only acts on the rectangle by stretching and squishing. You have one expanding eigenvalue (the one with absolute value grater than $1$) and one contracting eigenvalue (the one with absolute value less than 1 - note that it's negative so it also flips the rectangle).
So $L$ acts on $R_1$ by stretching it in the first eigendirection (as seen in the diagram) and squishing it in the other eigendirection, while also flipping it (which explains the new arrangement of the other rectangles around $R_1$).