If $r$ is a distance function both for $g$ and $g_{0}$ (something that in general need not be true), note how $\mathrm{Hess}_{0}\,r\,=\frac{g_{0}-(dr)^{2}}{r}$ in your flat metric $g_{0}$. I think you cannot replace $g_{0}$ with $g$ there. What I said in my answer below is still valid, I think: in general, the difference $\mathrm{Hess}\,{r}-\mathrm{Hess}_{0}\,r$ will depend on the specfic choice of a flat metric $g_{0}$.

Sorry, I didn't mean to render the answer invalid. I just wanted to make the question more meaningful.

@StefanKohl The general idea of my question has never changed. The first few versions are unclear and even the answerer did not see these versions. I just improved my question to make it concise and clear and even added my new thoughts. I made such a great effort, but you rolled back my question to the origin, it's really unfair.

@StefanKohl I have rolled back my question to the latest version without your permission. I think that the answer should be valid for all versions of my question as long as it is correct.

@Borromean, I wanted to keep this mathematical, but.. you changed your question many times until it transformed from a question about whether the difference of Hessians of a distance function is a multiplication of the Ricci to a question about asymptotic expansions. After I posted my answer, you and I had a discussion in the comments, but you deleted your own comments soon after. Still, I addressed your new question in the comment above, but you are choosing to ignore it and my answer. If you want something different now, then maybe you can open a new question instead of lingering here.

@MySheperd I never ignored your answer and I appreciate your attempt. I deleted my first guess that the difference is a multiplication of the Ricci tensor before you posted your answer and you had noticed that, so you edited your answer. Yes, I made two comments in your answer: the first one is about Laplacian but I said that the difference for Laplacian is O(1) which is wrong, so I deleted it. The second comment is important and maybe I should not delete it. I said that the space form is different from $M$ while you are considering two metrics both on $M$.

@MySheperd Your answer contains some correct facts and I was the first one to vote it up. But I think that your answer is not correct in general, at least it does not get the point of my question.

You did not document any changes you made to the original question. You started asking about asymptotic expansions only after I noted that the difference can be $o(r)$ in general, and only after we had a further discussion about this in the comments . Even if you get another idea from an answer, and realize your original question does not fit into exactly what you are looking for, it does not mean you need to change your question entirely to fit this new idea --- rendering previous answers invalid, as @StefanKohl put it. I think the right thing to do in this case is to open a new question.

In any event, I again bring to your attention that a distance function $r$ for a metric $g$ is in general not a distance function for a different metric $g_{0}$. Even if it is, and we write $g=dr^{2}+g_{r}$ and $g_{0}=dr^{2}+g^{0}_{r}$, note how $g_{r}$ is different than $g^{0}_{r}$. Thus, if the metric $g_{0}$ is flat, it is incorrect that $\mathrm{Hess}_{0}\,r=\frac{1}{r}g_{r}$, but rather $\mathrm{Hess}_{0}\,r=\frac{1}{r}\,g^{0}_{r}$. Again we find that $\mathrm{Hess}\,r-\mathrm{Hess}_{0}\,r$ depends on the exact choice of metrics $g$ and $g_{0}$, just like the original answer.

Now that you are so confirming that your answer is correct, why do you prevent the latest version of my question?

Hi Borromean, this is getting out of hand... what can we do to reach a compromise without keeping commenting on non-mathamtical subjects in the post?

Hi MySheperd, I do not want to waste time on non-mathematical subjects too.

Is it ok for you to discuss your question and my answer here and see if it satisfies you or not?

If it turns out you are not, then I suggest you make an Edit to your question, simply documenting how it differs from your original intent and the thought process you have put into it since first asked.

Sure, I can document the changes.

Thank you. But again, I must say I sincerely think that my answer demonstrates, in all versions, that any comparison of these Hessian will depend on the specific choice $g$ and $g_{0}$. These can be anything, so you will not have in the asymptotic expansion a concrete expression.

For example, in the example of my original answer, since the christoffel symbols are in relation to polar coordinates of $g$, the expansions will be indeed expressed in terms of curvature of $g$.

But this is not true in general, I can pick a different flat metric $g_{0}$ in which $r$ is not even a distance function, and then it will be an entirely different expression.

I think that the metrics are on different manifolds and the distance functions are also on different manifolds. When we compare the Hessians, we take the distance functions to be both $r$.

I added an example to my answer that demonstrates what I tried to say earlier. If the metrics are on different manifolds, how do you expect to compare the Hessians? Hessians are tensors defined over the tangent bundle of a manifold.