Closed form solution of linear regression via least squares using matrix derivatives
Matrix derivatives work a bit different than regular ones. The scalar parallel of $\nabla_{\theta} \theta^{T} X^{T}X\theta$ you make should be more like $\theta x^2 \theta$ which in turn is just $x^2 \theta^2$. (Note that I changed $\alpha$ for $x$ to avoid confusion with $X$.) You do not need to modify $\theta^{T} X^{T}X\theta$ so that the two $\theta$'s are beside each other, this actually is what you want.

Just like you usually have $\frac{d}{d\theta} (x^2 \theta^2) = 2 x^2 \theta$, then in matrix notation the rule is
$$\nabla_{\theta} \theta^{T} X^{T}X\theta = 2 X^{T}X \theta.$$

Hence, I feel he's not really "skipping steps", but applying a different step than the one you expected.