The Frisch-Waugh-Lovell Theorem: an exercise
What we know from FWL theorem, is that the regression

$$M_1y = M_1X_2\beta_2 + M_1u \tag{1}$$

will give the same estimates for $\beta_2$ as the full regression

$$y = X_1\beta_1 +X_2\beta_2 + u \tag{2}$$ 

where 

$$M_1 = I - P_1 = I - X_1(X_1'X_1)^{-1}X_1'$$ 

is the so-called annihilator or residual-maker matrix.  The estimator from $(1)$ is

$$\hat \beta_2 = (X_2'M_1X_2)^{-1}X_2'M_1y \tag{3}$$

So it boils down to examine whether the estimator from the specification 

$$P_1y = P_1X_2\beta_2 + w \tag{4}$$

which is

$$\tilde \beta_2 = (X_2'P_1X_2)^{-1}X_2'P_1y \tag{5} $$

will be the same as $\hat \beta_2$.

Well,

$$(2),(3) \implies \hat \beta_2 - \beta_2 = (X_2'M_1X_2)^{-1}M_1u \tag{6}$$

while 

$$ (2), (5) \implies \tilde \beta_2 -\beta_2 = (X_2'P_1X_2)^{-1}X_2'X_1\beta_1+ (X_2'P_1X_2)^{-1}X_2'P_1u \tag{7}$$

Given that $(6)$ and $(7)$ involve arbitrary exogenous quantities ($u, \beta_1$) I don't see how they could be equal, except by zero-probability chance.

Even if in our sample $X_1$ and $X_2$ are orthogonal (which would eliminate the first term in $(7)$ but which, with observational data, is a joke to even mention), then the two would be unbiased under strict exogeneity -but this is as far as similarities appear to go here.