The derivation of the score process is correct. To verify that the process is a Martingale, recall the [definition][1]. It becomes clear if substitute $W_{t+1}$ back into the equation $$ s_t(\theta \mid \textbf X) = \begin{bmatrix} (1 - \beta_0) \sum_{j=1}^t W_j \\ \sum_{j=1}^t W_j (X_{j-1} - \alpha_0) \end{bmatrix}. $$ Because $W_{t+1}$ are Normal with mean 0 and variance 1 (I'm assuming they're iid), then $$ E[s_{t+1} \mid s_t ] = s_t + E \left [ \begin{matrix} (1 - \beta_0) W_{t+1} \\ W_{t+1} (X_t - \alpha_0) \end{matrix} \middle | s_t \right ] = s_t $$ and we are finished. [1]: https://en.wikipedia.org/wiki/Martingale_(probability_theory)#Definitions

The derivation of the score process is correct. To verify that the process is a Martingale, recall the [definition][1]. It becomes clear that if we substitute $W_{t+1}$ back into the equation $$ s_t(\theta \mid \textbf X) = \begin{bmatrix} (1 - \beta_0) \sum_{j=1}^t W_j \\ \sum_{j=1}^t W_j (X_{j-1} - \alpha_0) \end{bmatrix}. $$ Because $W_{t+1}$ are Normal with mean 0 and variance 1 (I'm assuming they're iid), then $$ E[s_{t+1} \mid s_t ] = s_t + E \left [ \begin{matrix} (1 - \beta_0) W_{t+1} \\ W_{t+1} (X_t - \alpha_0) \end{matrix} \middle | s_t \right ] = s_t $$ and we are finished. [1]: https://en.wikipedia.org/wiki/Martingale_(probability_theory)#Definitions