Suppose we have some self adjoint operator, given by either a matrix $$\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$$ or a function $f \longmapsto xf$ on $L^2[-1,1]$. Is there a quick way of decomposing these self adjoint operators into the difference of positive operators?

Compose the operator with the projections on the positive and negative part of the spectrum and take the difference. For your matrix $A$, you need to find the eigenvalues which are the roots of

$$\lambda^2 - \operatorname{tr}(A)\lambda + \det(A) = \lambda^2 - 2\lambda -3 = (\lambda + 1)(\lambda - 3).$$

An eigenvector associated to $\lambda = 3$ is $v_1 = (1,1)^T$ and an eigenvector associated to $\lambda = -1$ is $v_2 = (-1,1)^T$. The orthogonal projection onto $\operatorname{span} \{ v_1 \}$ is given by

$$P_1 = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}$$

and the orthogonal projection onto $\operatorname{span} \{ v_2 \} = \operatorname{span} \{ v_1 \}^{\perp}$ is given by

$$P_2 = I - P = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix}.$$

Finally, we have

$$A = A(P_1 + P_2) = AP_1 - (-AP_2) = \begin{pmatrix} \frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & \frac{3}{2} \end{pmatrix} - \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\end{pmatrix}$$

where $AP_1$ is positive (with spectrum $\{ 0, 3 \}$) and $-AP_2$ is positive (with spectrum $\{ 0, 1 \}$).

For the operator $(M(f))(x) = x f(x)$ on $L^2([-1,1])$ the situation is easier because it is already "diagonal". Just define

$$g_1(x) = \begin{cases} x & x \geq 0, \\ 0 & x < 0 \end{cases}$$

and let $P_1(f)(x) := g_1(x) f(x)$. Then $P_1$ is positive (with spectrum $[0,1])$ and $M = P_1 - (P_1 - M)$ where $P_1 - M$ is also positive.