I believe that this reasoning is incorrect. Solving the above equation without a regularizer (assuming that $A^TA$ is invertible) would lead to a large condition number.

Tikhonov regularization is directly used to combat that as is shown in the following Lemma: $$ d_{ii}=\frac{1}{\sigma_i + \lambda}. $$ So the inverted singular values are stabilized using the regularization parameter $\lambda$, removing the issue of small singular values being inverted and blowing up the whole prediction.

Or perhaps I am misunderstanding the wording here.

Shouldn't the n-th derivative evaluate to

$$ g^{(n)}(\hat{q})=\frac{n!}{(1-\hat{q})^{n+1}} $$

Because otherwise the Taylor polynomial would evaluate to

$$ T_g(\hat q; 0)=\sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!}(\hat q - 0)^n=\sum_{n=0}^\infty \frac{n}{n!}\hat{q}^n $$

This should say 2D