$m_p=\{f\in \mathcal{O}_{V,p}| f(p)=0\}$, ideal of $p$ in the local ring. What is $m_p/m_p^2$? In Section 6.8 of Undergraduate Algebraic Geometry by Reid, the author proved the following Theorem:

There is a natural isomorphism of vector spaces $(T_pV)^*\cong m_p/m_p^2$ where $^*$ denotes the dual of a vector space.

Here $T_pV$ is the tangent space to $V$ at $p$, where $V$ is a variety in $\mathbb{A}^n$. $m_p$ is the ideal of $p$ in $k[V]$. For simplicity, $p$ is assumed to be $(0,\dots,0)$. 
$M_p$ is the ideal $(x_1,\dots,x_n)$ in $k[x_1,\dots,x_n]$. I understand the part showing 
$$M_p/M_p^2\cong (k^n)^*$$
He then introduced a restriction map $(k_n)^*\rightarrow (T_pV)^*$ and states that
$$m_p/m_p^2=M_p/(M_p^2+I(V))\cong (T_pV)^*$$
this is where I got lost. I understand why the quotient $M_p/M_p^2$ is the vector space of linear function since $M_p^2$ is the ideal generated by the second order functions. But I am not sure what $m_p/m_p^2$ is and why it is equal to $M_p/(M_p^2+I(V))$.
Thank you for your help!

$m_p$ is the maximal (irrelevant) ideal of $K[X_1,\dots,X_n]/I(V)$, that is, the ideal generated by the images of $X_1,\dots,X_n$ hence $m_p=M_P/I(V)$. Then $m_p^2=(M_P^2+I(V))/I(V)$. Now it's clear why $$m_p/m_p^2=M_P/(M_P^2+I(V)).$$
