It's convenient to define $u_1=y_1+y_2$ and $u_2=y_1-y_2$. Then we find (removing the explicit use of the indicator function) $u_1''=\left\{ \begin{array}{ll} -2\frac{A}{R}r, & |r|>R;\\ B(1+R)u_1 - 2\frac{A}{R}r, & |r|\leq R, \end{array} \right. $ and $u_2''=\left\{ \begin{array}{ll} 0, & |r|>R;\\ B(R-1)u_2, & |r|\leq R. \end{array} \right. $
For $|r|>R$, we can write down general solutions by integrating twice.
For $|r|\leq R$, we obtain first general solutions of the homogeneous equations $u_1''=B(1+R)u_1$ and $u_2''=B(R-1)u_2$. Assuming that $R$ is also a constant, these are second order linear constant coefficient equations, so we can write down the required general solutions. For $u_2$, we're done, and for $u_1$, we look for a particular solution using the method of undetermined coeffients (that is, making an educated guess).
Assuming that the solutions should be twice-differentiable for all $r$, we then need to match the interior and exterior solutions (and their first and second derivatives) at $|r|=R$ to ensure this. This would involve choosing the constants of integration appropriately.
For a system with $n$ equations of the form $y_i''=\alpha(r)+\beta(\rho y_i+\sum_{i\neq j}y_j),$ we can write $y''=a(r) + \beta(\rho I+J)y,$ where $y\in\mathbb{R}^n$, $a$ is the column vector with every entry equal to $\alpha$, $I$ is the $n\times n$ identity matrix and $J$ is the matrix with ones everywhere except the diagonal, which has zeroes. $\beta$ and $\rho$ are constants corresponding to $B,R$ respectively. Let $M=\rho I +J$. This has eigenvalues $\rho+n-1$ (with eigenvector $(1,1,\dots,1)^T$) and $\rho-1$. The eigenvectors with eigenvalue $\rho-1$ are of the form $(y_1,y_2,\dots,y_n)^T$ with $y_1+y_2+\cdots+y_n=0.$ Thus $\rho-1$ has an $(n-1)$-dimensional eigenspace. Hence $M$ has a complete set of eigenvectors, and so can be diagonalized: there exists a non-singular matrix $S$ such that $S^{-1}MS=\Lambda=\rm{diag}(\rho+n-1,\rho-1,\dots,\rho-1).$
Now define $u\in\mathbb{R}^n$ by $y=Su$. Since $S$ is a constant matrix, we can follow the definitions above to get $u''=S^{-1}a+\beta\Lambda u.$ Since $\Lambda$ is diagonal, this provides decoupled second order linear equations for the components of $u$.