According to lemma 4.2.3 of Qing Liu's book on Algebraic Geometry, the canonical homomorphism $\mathfrak{m}/\mathfrak{m}^2 \to \mathfrak{m}A_\mathfrak{m}/ \mathfrak{m}^2 A_\mathfrak{m}$ is an isomorphism of $A/\mathfrak{m}$ vector spaces.
To show this, the book proves that $\varphi: A/\mathfrak{m} \to A_\mathfrak{m}/\mathfrak{m}A_\mathfrak{m}$ is an isomorphism and says it suffices to tensor this map by $\otimes_A \mathfrak{m}$. I don't know why this is enough, wouldn't it be required that $\mathfrak{m}$ is a flat $A$-module?
$\newcommand{\m}{\mathfrak m}$ You have an exact sequence, regardless of flatness: $\m\otimes_A \m \to \m\to \m\otimes_A A/\m\to 0$ (note that flatness would add a leftmost $0$ but we don't need it)
Moreover, the image of $\m\otimes_A\m\to \m$ is clearly $\m^2$. This is not to say that $\m\otimes_A\m \cong \m^2$ which would indeed require flatness of $\m$. It follows that $\m/\m^2\cong \m\otimes_A A/\m$
Similarly, one has an exact sequence $\m\otimes_A \m A_\m \to \m A_\m\to \m\otimes_A A_\m/\m A_\m\to 0$, and we now have to check that the image of $\m\otimes_A \m A_\m \to \m A_\m$ is $\m^2 A_\m$, but this is clear too; and so similarly $\m A_\m/\m^2A_\m\cong \m\otimes_A A_\m/\m A_\m$