$\frac{\mathfrak{m}_a}{\mathfrak{m}_a^2}\cong \text{Coker}\left (\text{J}\varphi \left ( a \right ) \right )$ Let $k$ be an algebraically closed field and let $\varphi :k^n\to k^r$ be a map such that there exist $f_1,\cdots ,f_r\in k[x_1,\cdots ,x_n]$ such that for every $a\in k^n$, $\varphi (a)=(f_1(a),\cdots ,f_r(a))$.
For $a\in k^n$ such that $f_i(a)=0$ for every $i$, define $\text{J}\varphi \left ( a \right )\in k^{n\times r}$ as $\left [\displaystyle \frac{\partial f_j}{\partial x_i}\left (a \right )\right ]_{ij}$. Let $\mathfrak{m}_a\subset k[x_1,\cdots ,x_n]/\langle f_1, \dots, f_r \rangle $ be the maximal ideal $\langle x_1-a_1,\cdots ,x_n-a_n \rangle$.
Prove that $\frac{\mathfrak{m}_a}{\mathfrak{m}_a^2}\cong \text{Coker}\left (\text{J}\varphi \left ( a \right ) \right )$ as $k$-vector spaces.
 A: What is $m_a$ exactly? It's the ideal containing all regular functions on the variety $(f_1 =  \dots = f_r = 0) \subset A^n$ that vanish at $a$. In other words, it contains all regular functions such that the constant term of their Taylor expansion around $a$ vanish.
And what is $m_a^2$? It's the ideal containing all the regular functions whose Taylor series around $a$ have vanishing constant AND linear terms.
So $m_a/m_a^2$ is the vector space of "linear truncations of regular functions on $(f_1 =  \dots = f_r = 0) \in \subset A^n$ vanishing at $a$". In other words $m_a / m_a^2$ is spanned by $x_1 - a_1, \dots , x_n - a_n$ as a $k$-vector space.
Except...
$x_1 - a_1, \dots , x_n - a_n$ are not linearly independent in $m_a/m_a^2$! This is because we are quotienting by the ideal $(f_1, \dots , f_r)$. So really, $m_a / m_a^2$ is the vector space generated by $x_1 - a_1, \dots , x_n - a_n$, quotiented by the subspace generated by the linear terms of the Taylor expansions of $f_1, \dots , f_r$.
The Taylor expansion of $f_j(x)$ is
$$ f_j(x) = \partial_1 f_j(a) (x_1 - a_1) + \dots + \partial_n f_j(a) (x_n - a_n)+ {\rm \ higher \  order \ }. $$
(Remember that $f_j(a) = 0$.)
So $m_a / m_a^2$ is the vector space generated by $x_1 - a_1, \dots , x_n - a_n$, quotiented by the subspace generated by linear combinations of $$\partial_1 f_1(a) (x_1 - a_1) + \dots + \partial_n f_1(a) (x_n - a_n), \\ \vdots \\ \partial_1 f_r(a) (x_1 - a_1) + \dots + \partial_n f_r(a) (x_n - a_n).$$
To finish off, consider the linear map $k^n \to m_a / m_a^2$, mapping
$$ (c_1, \dots c_n) \mapsto c_1 (x_1 - a_1) + \dots + c_n (x_n - a_n).$$
It is surjective, and its kernel is spanned by the vectors
$$(\partial_1 f_1(a) , \dots , \partial_n f_1(a) ), \  \dots \  , (\partial_1 f_r(a) , \dots , \partial_n f_r(a) ).$$
This kernel is clearly ${\rm Im \ }J (\varphi (a))$. So ${\rm Coker \ J(\varphi(a))}  = m_a / m_a^2$.
Finally, as you pointed out in your comment, $m_a / m_a^2 \cong m_aA_a / (m_a A_a)^2 $. So we've shown ${\rm Coker \ J (\varphi(a))}  = m_a A_a / (m_a A_a)^2$. Both of these vector spaces are representations of the cotangent space at $a$. The expression ${\rm Coker \ J (\varphi(a))}$ is closer to your intuitive notion of the cotangent space, but the expression $m_aA_a / (m_a A_a)^2$ has the advantage that it is intrinsic to the local ring at the point of the variety.
