Why is it that $M_{X_2} M_X = M_X$ for two different residual makers? I can't seem to figure this out algebraically, or intuitively.
Same with a result like $M_X M_X = M_X$ which I know is the idempotent property for a residual maker.
If I think of $M_X X = 0$, this makes sense to me, because it's saying the vector of least-squares residuals of regressing X on itself is 0 - because X perfectly predicts itself.
But what's the interpretation when we have two residual makers? What exactly is the regression?
 A: If you take some vector $y$ and regress on $X$ you get residuals $\hat \epsilon = M_Xy$ which is why $M_X$ is sometimes referred to as the "residual maker matrix". Let assume you decide to take these residuals and regress them on $X_2$ where $X = [X_1 \ X_2]$ then the identity that $M_{X_2}M_X=M_X$ for $X = [X_1 \ X_2]$ tells you that the residuals $\hat u$ from that regression will be $\hat u=M_{X_2}\hat \epsilon =M_{X_2} M_Xy = M_Xy = \hat \epsilon$ the same. 
You could say that the first regression takes all the variation in $y$ that systematically varies with $X$ and puts the unsystematic variation in $\hat \epsilon$. Since $X_2$ is included in $X$ there is no variation left in these residuals that systematically varies with $X_2$. 
Proof
To show that $M_{X_2}M_X=M_X$ for $X = [X_1 \ X_2]$. 
It should makes sense to you that $$P_X X = X,$$ which can be read as for any column in $X$ the predicted values from a regression of the column unto $X$ is the column itself. 
And by partioning this implies that $P_X [X_1 \ X_2] = [X_1 \ X_2]$ hence $$P_XX_2 = X_2,$$
stating simply that since it holds for any column of $X$ it must hold for some selection of the columns in $X$ the selection being $X_2$.
Using this it is easy to show that $P_{X_2}P_X=P_{X_2}$. Consider that
$$P_{X_2}P_X=X_2(X_2^\top X_2)^{-1}X_2^{\top}P_X$$
and notice that the final part $X_2^{\top}P_X$ is simply $(P_XX_2)^\top$ due to symmetry of $P_X$. But you know from above that $P_XX_2=X_2$ so $(P_XX_2)^\top = X_2^\top$. It follows then that
$$P_{X_2}P_X=X_2(X_2^\top X_2)^{-1}X_2^{\top}P_X = X_2(X_2^\top X_2)^{-1}X_2^{\top} =P_{X_2}$$
Finally consider that
$$M_{X_2}M_X = (I-P_{X_{2}})(I-P_X) = I  -P_{X} - P_{X_2}+ P_{X_2}P_X =  I  -P_{X} = M_X$$
using from above that  $P_{X_2}P_X=P_{X_2}$.
