Note that the differential $d\theta$ is simply the gradient of the path defined above. In the case of $\wp$ we see that the path is given by $\theta \mapsto (\wp(\theta), \wp'(\theta)) = (x,y)$, hence the gradient is given by $(\wp'(\theta), \wp''(\theta))$. Next, we derive the functional equation $$ (\wp'(\theta))^2 = 4(\wp(\theta))^3 - g_2 \wp(\theta) - g_3 $$ to obtain $$ 2 \wp'(\theta) \wp''(\theta) = 12 (\wp(\theta))^2 \wp'(\theta) - g_2 \wp'(\theta) $$ and hence $$ \wp''(\theta) = 6(\wp(\theta))^2 - \frac{g_2}{2} $$ Recalling that $\wp(\theta) = x$, we see that $\wp''(\theta) = 6x^2-\frac{g_2}{2}$, hence we may write $$ d\theta = ydx + (6x^2 - \frac{g_2}{2})dy $$
Hope that this fully answers your question.