This is Construction 11.34 from Gathmann's Notes on Commutative Algebra.
$X\subset A^n$ is a variety with coordinate ring $A(X)=k[x_1,\dots, x_n]/I(X)$. Let $d:k[x_1,\dots, x_n]\to k[y_1,\dots,y_n]$ by $f\to \sum_i\frac{\partial f}{\partial x_i}(a)y_i$ where $a=(a_1,\dots, a_i)\in X$ and $y_i=x_i-a_i$.
Define $T_aX=V(L)\subset K^n$ where $L=\{dg:g\in I(X)\}$. Let $I(a)$ be the corresponding maximal ideal of $a$ in $A(X)$.
$\phi: I(a)\to Hom_K(T_aX,K)$ by $\bar{f}\in I(a)\to df\vert_{T_aX}$. It can be checked the map is well defined.
It is clear that $L$ is a m-dimensional $K$ vector subspace $\subset (y_1,\dots y_n)$ for some $m\leq n$. The zero locus, $V(T_aX)=\{(y_1,\dots, y_n)\in K^n\vert (df)(y_1,\dots, y_n)=0, f\in I(X)\}$ is of dimension $n-m$.
I do not get the idea of the following though I feel rank nullity theorem is used somewhere.
"Hence the space of all linear forms vanishing on $T_aX$ has dimension $m$ again and clearly contains $L$, and thus must be equal to $L$. As $df$ lies in this space, we conclude that $df=dg$ for some $g\in I(X)$."
I tried to make sense of the statement by computing $A(X)=C[x_1,x_2]/(x_2^2-x_1)$ with $a=(0,0)$. $T_aX=V(\{df\vert f\in (x_2^2-x_1)\})$. So $f=h(x_2^2-x_1)$ with $df=d(h(x_2^2-x_1))=h(0,0)(-x_1)$, I get $T_aX=V(x_1)$. $\phi:I((0,0))\to Hom_K(T_{(0,0)}X,K)$. $L=\{-h(0,0)x_1\}$ which is a linear vector space. So $\phi(f)=df|_{(0,x_2)}=0$ for $df$ containing only non-trivial $x_1$ component which is exactly what the statement says.
How should I formally understand the statement in the quotation?
This is basically what I have in my mind. All vector spaces here are finite dimensional. Suppose $L\subset (y_1,\dots, y_n)$ is an $m-$dimensional vector space. Then $T_aX$ is a $n-m$ dimensional vector space by thinking basis of $L$ as linearly independent relation enforced on $y_i$. Where $df|_{T_aX}=0$ says $df$ is perpendicular to $T_aX$. There is only 1 vector space $L$ orthogonal to $T_aX$. So I conclude that $df|_{T_aX}=0$ implies $df\in L$. And the rest follows.