Let $f:[0,1]\times [0,1] \to \mathbf{R}$ be infinitely differentiable, non-negative, and have finitely many zeros. Additionally, let $(a,b)\in [0,1]\times [0,1]$ be an arbitrary zero of $f$. Let $$ \widetilde{f}(x,\xi) = \sum_{0\leq i+j< \infty}\frac{\partial^{i+j}f(a,b)}{\partial x^i \partial \xi^j}\frac{(x-a)^i(\xi-b)^j}{i!j!} $$ denote the Taylor series of $f$ centered about the point $(a,b)$, which converges for points in the ball $B_{\delta}$ with radius $\delta>0$ and for such points we also have that $f(x,\xi) = \widetilde{f}(x,\xi)$. Now, further suppose that the first nonzero derivative of $f$ is $\frac{\partial^2}{\partial \xi^2}f(a,b)$. Additionally, there exists an $n\geq 1$ such that $\frac{\partial^n}{\partial x^n}f(a,b) \ne 0$.
Now, do there exist positive constants $c,k, \widetilde{\delta}$ such that for $(x,\xi)\in B_{\widetilde{\delta}}$, $$ |f(x,y)|\geq c ((x-a)^2+(\xi-b)^2)^{\frac{k}{2}}. $$
I don’t have an answer, but I will write an example that completely contradicts what I wrote in the comments.
Consider $$ f(x,y)=x^6+y^2-2x^3y+x^{100}+y^{100}. $$
Then, $f$ has an isolated zero at $(0,0)$, and satisfies your hypotheses with $n=6$, but when you truncate the Taylor expansion to any order between $6$ and $100$, you get a function with zeroes all along the curve $\{y=x^3\}$. So, reasoning by approximating $f$ with a truncated Taylor series, and simply reducing the statement to the case where $f$ is a polynomial, is not so trivial. Dumb me.
Remark 1. You can see that in this case the bound from below holds with $k=100$, although not with $k=n$. So $k$ can be completely unrelated from $n$.
Remark 2. You can also see that the statement is not true if $f$ is not analytic close to the zero. Simply modify my example slightly: the function $$ g(x,y)=x^6+y^2-2x^3y+e^{-1/(x^2+y^2)} $$ satisfies all your hypotheses (except that of being equal to its Taylor series around $(0,0)$), but does not have the lower bound you want for any $k>0$.