Decomposition of an additive functional into a Martingale part and other
Using the given formulas (inserting the expression for the $X$'s in the sum), we arrive at

$$Y_t = (1-\beta_0)\alpha_0\cdot t + \beta_0\sum_{j=0}^{t-1}X_j +\sum_{j=1}^{t}W_j $$

We do not need a theorem to obtain that, given the description of the problem.

Manipulating, 

$$Y_t= (1-\beta_0)\alpha_0\cdot t  - \beta_0(X_t-X_0) + \beta_0Y_{t}  +\sum_{j=1}^{t}W_j$$

$$\implies Y_t = \alpha_0t +\frac{1}{1-\beta_0}\sum_{j=1}^{t}W_j-\frac{\beta_0}{1-\beta_0}(X_t-X_0)$$

and we want to match this with 

$$Y_t = r_1 t + M_t - r_2 (X_t - X_0)$$

We immediately get that we must have  $r_1 =  \alpha_0$, $r_2 = \beta_0/(1-\beta_0)$, and 
$$M_t = \frac{1}{1-\beta_0}\sum_{j=1}^{t}W_j$$

Is $M_t$ as obtained a martingale? We have

$$E[M_t\mid \sigma(t-1,t-2,..)] = \\=\frac{1}{1-\beta_0}E[W_t\mid \sigma(t-1,t-2,..)] + \frac{1}{1-\beta_0}\sum_{j=1}^{t-1}W_j$$

$$= \frac{1}{1-\beta_0}E[W_t\mid \sigma(t-1,t-2,..)] + M_{t-1}$$

So if $E[W_t\mid \sigma(t-1,t-2,..)] = 0$, namely, if the original disturbance is mean-independent from the past (in a more fancy term, if $\{W_t\}$ is a "Martingale difference"), then $M_t$ is a martingale and we have our mapping.