It turns out that this statement can be rephrased as: The Minkowski-Bouligand dimension of $M$ is equal to $k$ (this is immediate from the definition). Another word for this is the box dimension, and we will write it as $\mathrm{dim}_\mathrm{B}$. Here's a proof sketch that $\mathrm{dim}_\mathrm{B}(M^k)=k$.

By the inverse function theorem and compactness, we can write $M$ as a collection of finitely many smooth graphs over its tangent spaces. More precisely, there exists a $\rho>0$, points $x_1,\dotsc, x_N\in M$, smooth functions $u_1,\dotsc,u_N$ with $u_i:B_{\rho}(x_i)\cap T_{x_i}M\to\Bbb R^{d-k}$ and $\mathrm{Lip}(u_i)\le L$ such that
$$M=\bigcup_{i=1}^N\mathrm{graph}(u_i).$$
It suffices to check the claim for each graph.

So let $B$ now be the $k$-dimensional unit ball and $u:B\to\Bbb R^{d-k}$ a Lipschitz function. It is clear by considering grids of spacing $\varepsilon$ that $\mathrm{dim}_\mathrm{B}(B)=k$. Let $\mathcal N_B(\varepsilon)$ be the covering number of $B$. Then $\mathcal N_{\mathrm{graph}(u)}(L\varepsilon)\le \mathcal N_B(\varepsilon)$. Therefore, $\mathrm{dim}_\mathrm{B}(\mathrm{graph}(u))\le \mathrm{dim}_\mathrm{B}(B).$

For the other inequality, we may assume that our graphical parametrization is in fact bi-Lipschitz by making the original balls smaller (but still having finitely many). That is, we may assume that $B_{\rho}(x_i)\cap T_{x_i}M$ is a Lipschitz graph over a subset of $B_{\rho}(x_i)\cap M$. This gives $\mathrm{dim}_\mathrm{B}(\mathrm{graph}(u))\ge \mathrm{dim}_\mathrm{B}(B)$, which completes the proof.