Suppose we have a discrete map of the form
$$x_{n+1} = f(x_n) \qquad x \in \Bbb R$$
where $f:\Bbb R\rightarrow \Bbb R$. Then by definition, a $2$-cycle is a periodic orbit $\{y_1,y_2\}$ with $y_1 \neq y_2$ such that
$$f(y_1) = y_2 \qquad f(y_2) = y_1$$
To check the stability of this orbit, we compute the quantity $|f'(y_1)f'(y_2)|$ to see whether it is larger, equal to, or less than $1$.
So my question is, if we have a 2D map of the form
$$\begin{pmatrix} x_{n+1} \\ y_{n+1} \end{pmatrix} = \mathbf f \begin{pmatrix} x_{n} \\ y_{n} \end{pmatrix} \qquad x_n, y_n \in \Bbb R$$
where $\mathbf f:\Bbb R^2 \rightarrow \Bbb R^2$, then how do we determine the stability of a 2-cycle?
To avoid confusion about the use of the $y$ variables, let the stable cycle in 2-D be at $(X_a,Y_a)$ and $(X_b,Y_b)$.
In principle, you would let $x_n = X_a + \epsilon_x, x_n = Y_a + \epsilon_y$, then move forward two steps discarding all terms higher than linear in teh $\epsilon$s. This cives a linear transformation between $(\epsilon_x^(n), (\epsilon_y^(n))$ and $(\epsilon_x^(n+2), (\epsilon_y^(n+2))$. Diagonalize the matrix of that linear transformation. If the absolute values of the eigenvalues are all $<1$ then the point is stable. If any eigenvalue is greater than $1$, then the point is unstable.
As a practical matter, you can do the perturbation theory quickly as follows:
Form the two matrices $$ M_a = \pmatrix{ \left.\frac{\partial f_x}{\partial x}\right|_{X_a,Y_a} &\left.\frac{\partial f_y}{\partial x}\right|_{X_a,Y_a} \\ \left.\frac{\partial f_x}{\partial y}\right|_{X_a,Y_a} &\left.\frac{\partial f_y}{\partial y}\right|_{X_a,Y_a} } \\ M_b = \pmatrix{ \left.\frac{\partial f_x}{\partial x}\right|_{X_b,Y_b} &\left.\frac{\partial f_y}{\partial x}\right|_{X_b,Y_b} \\ \left.\frac{\partial f_x}{\partial y}\right|_{X_b,Y_b} &\left.\frac{\partial f_y}{\partial y}\right|_{X_b,Y_b} } $$ and then the matrix product $ M = M_aM_b $ which is the linearized transition matrix mapping the vector from point $a$ at time $n$ to point $a$ at time $n+2$. This is called the Jacobian matrix. If the absolute values of the eigenvalues of $M$ are all $<1$ then the point is stable. If any eigenvalue is greater than $1$, then the point is unstable.
What happens in the boundary case, where one (or both) eigenvalue is $\pm 1$?
That can get quite interesting as second (and perhaps higher) derivatives come into play. The issue is that the cross second derivatives can mean that the eigenvectors of the second derivative matrix (the Hessian matrix) need not align with the eigenvectors of the Jacobean matrix. So you could have a case where one eigenvalue of the Jacobian is 1, the second derivative is positive (hinting at instability), but the effect of the second derivatives cause a mix-in of the stable direction, with the net effect that the perturbed trajectory does spiral in to the (stable) point.