Let $Y$ be an affine variety in $\mathbb{A}^n_k$ and $\mathfrak{i}$ its corresponding ideal. We use the notation $A(Y) = k[x_1,...,x_n]/\mathfrak{i}$ for the coordinate ring of $Y$. Pick a point $p\in Y$, we use the notation $\mathcal{O}_p(Y)$ to be the collection of all pairs $(U,f)$ where $U$ is an open subset of $Y$, containing $p$, modulo the relation that $(U,f)\sim (V,g)$ iff $f=g$ on $U\cap V$. This is the ring of regular functions of $Y$ at the point $p$. We also introduce the notation $\mathfrak{p} = (x_1-p_1,...,x_n-p_n)$ as the maximal ideal corresponding to $p = (p_1,...,p_n)$. Note, the ring $A(Y)$ has $\mathfrak{i}' = \mathfrak{p}/\mathfrak{i}$ for its maximal ideal.
We know that $\mathcal{O}_p(Y) \simeq A(Y)_{\mathfrak{i}'}$, an isomorphism of vector spaces over $k$. Let us denote $\mathfrak{m}$ for the maximal ideal of $\mathcal{O}_p(Y)$, that is, the ideal of rational functions which vanish at $p$. It therefore follows that, $$ \mathfrak{m}/\mathfrak{m}^2 \simeq \frac{ S^{-1} \mathfrak{i}'}{ \left(S^{-1} \mathfrak{i}'\right)^2} $$ Where $S$ is the localizing set i.e. $S$ is the complement of $\mathfrak{i}'$. How do we conclude that, $$ \frac{ S^{-1} \mathfrak{i}'}{ \left(S^{-1} \mathfrak{i}'\right)^2}\simeq \frac{\mathfrak{p}}{\mathfrak{i} + \mathfrak{p}^2} $$
Basically you have a ring $R$, a maximal ideal $\mathfrak p$ and $\mathfrak i\subset\mathfrak p$ an ideal. Then in $R/\mathfrak i$ you consider the maximal ideal $\mathfrak p/\mathfrak i$. The localization $(R/\mathfrak i)_{\mathfrak p/\mathfrak i}$ is isomorphic to $R_{\mathfrak p}/\mathfrak iR_{\mathfrak p}$ and $$\displaystyle\frac{\mathfrak pR_{\mathfrak p}/\mathfrak iR_{\mathfrak p}}{(\mathfrak pR_{\mathfrak p}/\mathfrak iR_{\mathfrak p})^2}=\frac{\mathfrak pR_{\mathfrak p}/\mathfrak iR_{\mathfrak p}}{(\mathfrak p^2R_{\mathfrak p}+\mathfrak iR_{\mathfrak p})/\mathfrak iR_{\mathfrak p}}\simeq\frac{\mathfrak pR_{\mathfrak p}}{\mathfrak p^2R_{\mathfrak p}+\mathfrak iR_{\mathfrak p}}\simeq\left(\frac{\mathfrak p}{\mathfrak p^2+\mathfrak i}\right)_{\mathfrak p}.$$
The last step is to show that $$\displaystyle\left(\frac{\mathfrak p}{\mathfrak p^2+\mathfrak i}\right)_{\mathfrak p}\simeq\frac{\mathfrak p}{\mathfrak p^2+\mathfrak i}.$$ In order to prove this consider the canonical morphism $$\displaystyle\phi:\frac{\mathfrak p}{\mathfrak p^2+\mathfrak i}\to\left(\frac{\mathfrak p}{\mathfrak p^2+\mathfrak i}\right)_{\mathfrak p}$$ It's easily seen that $\phi$ is injective. (Use that $\mathfrak p^2+\mathfrak i$ is $\mathfrak p$-primary.) For surjectivity we have to show the following: given $a\in\mathfrak p$ and $s\in R-\mathfrak p$ there is $b\in\mathfrak p$ such that $a-sb\in\mathfrak p^2+\mathfrak i$. Since $\mathfrak p/\mathfrak p^2$ is an $R/\mathfrak p$-vector space we can consider $\bar b\in\mathfrak p/\mathfrak p^2$ such that $\bar b=\hat s^{-1}\bar a$, hence $\hat s\bar b=\bar a$, so $a-sb\in\mathfrak p^2$.