Two R² values, one joint fit

Idea. The lower end comes from nesting: the joint model contains each single model. The upper end comes from geometry: the two given angles do not stop $y$ from lying in the plane spanned by $x_1$ and $x_2$.

Set-up. Centre all three variables and view them as vectors. With one regressor, $R^2$ equals the squared correlation, which is the squared cosine of the angle between the two vectors. So $${\cos }^2{\theta }_1=0.15,{\cos }^2{\theta }_2=0.2,$$ where ${\theta }_i$ is the angle between $y$ and $x_i$. The joint $R^2$ is the squared cosine of the angle between $y$ and the plane spanned by $x_1$ and $x_2$.

Lower end. The joint fit can always set the coefficient of $x_1$ to zero, so its $R^2$ is at least $0.2$. This is reached when $x_1$ carries no information about $y$ beyond what $x_2$ already gives, for example $x_1=a{x}_2+z$ with $z$ uncorrelated with both $y$ and $x_2$, and $a$ chosen so that $\mathrm{corr}(y,x_1{)}^2=0.15$. So $m=0.2$.

Upper end. Place $x_1$ and $x_2$ in a plane that contains $y$, at angles ${\theta }_1$ and ${\theta }_2$ from it, on opposite sides. Both given correlations hold, $x_1$ and $x_2$ are not collinear, and $y$ lies in their span, so the joint fit is perfect. So $M=1$.

Answer. $$\frac{M}{m}=\frac{1}{0.2}=5.$$

The usual trap. Adding the two values gives $0.35$. That is the joint $R^2$ only when $x_1$ and $x_2$ are uncorrelated. In general, with $r_1,r_2$ the correlations of $y$ with $x_1,x_2$ and $\rho =\mathrm{corr}(x_1,x_2)$, $$R^2=\frac{r_1^2+r_2^2-2r_1{r}_2\rho }{1-{\rho }^2},$$ and as $\rho$ ranges over the values that keep the correlation matrix valid, this sweeps the whole interval $[0.2,1]$.