Can we interpret the "extent of reaction" as the number of reactions that happened?
If I understand this correctly you want to measure $\ce A$ to infer the extent of reaction. I think you should have started with mathmatically defining the extend of reaction:

$$\xi(t) = \left[\frac{n_i(\ce A) -n_t(\ce A)}{n_i(\ce A)}\right]\tag 1$$

Where $n_t$ is the concentration at a given time $t$. Here $n_i(\ce A) -n_t(\ce A)$ gives the consumed concentration of $\ce A$ after time $t$ which is normalized by the initial concentration to yield  an extent of conversion. 


  Assume $\mathrm{2\ a}$: If the stoichometric constants are normalized $({\alpha} + {\beta}= 1)$ to be taken as the initial concentrations then we can substitute: $\alpha = n_i(\ce A)$ for example $$\ce{0.33A + 0.67B -&gt; 0.5C + 0.5D}$$
  $$\xi(t) = \left[\frac{\alpha -n_t(\ce A)}{\alpha}\right]\tag {2 a}$$
  otherwise:
  $$\left(\frac{\alpha}{\alpha + \beta} + \frac{\beta}{\alpha + \beta} = 1\right) \implies \frac{\alpha}{\alpha + \beta} = n_i(\ce A)\tag {2 b}$$


Which we can rearrange to :

$$\xi(t) = \left[\frac{\alpha -n_t(\ce A)}{\alpha}\right] \implies \alpha-n_t(\ce A) = \alpha\ \xi(t) \tag{3} $$

Subtracting $\alpha$ and then negating gives:

$$n_t(\ce A) = \alpha - \alpha\ \xi(t) = n_i(\ce A) - \alpha\ \xi(t) \tag{4 a}$$
Which would make you correct, or in the case of assumption $\mathrm{2\ b}$:
$$n_t(\ce A) = n_i(\ce A) - \frac{\alpha}{\alpha + \beta}\ \xi(t) \tag{4 b}$$
Which would make you almost correct.