I am learning Linear Algebraic Groups without enough knowledge on Algebraic Groups. I see the definition for the differential of a morphism on page 42 of James Humphreys' Linear Algebraic Groups (GTM 21):
Let $\phi: X \rightarrow Y$ be a morphism of (irreducible) varieties. If $x \in X$, $y = \phi(x)$, then $\phi^*$ maps $(\mathcal{o}_y, \mathcal{m}_y)$ into $(\mathcal{o}_x, \mathcal{m}_x)$. By composition with $\phi^*$, a linear function x on $\mathcal{m}_x/ \mathcal{m}_x^2$ therefore induces a linear function on $d \phi _x($x$)$ on $\mathcal{m}_y/ \mathcal{m}_y^2$. The resulting map $d \phi_x: \mathcal{T}(X)_x \rightarrow \mathcal{T}(Y)_y$ is evidently $K$-linear. We call it the differential of $\phi$ at $x$.
Here, $K$ is an algebraically closed field. $\phi^*: K[Y] \rightarrow K[X]$, $f \mapsto f \circ \phi$. $\mathcal{o}_x$ (resp. $\mathcal{o}_y$) represents the local ring of regular functions at $x$ (resp. $y$) with maximal ideal $\mathcal{m}_x$ (resp. $\mathcal{m}_y$).
I think, if $\phi^*$ maps $(\mathcal{o}_y, \mathcal{m}_y)$ into $(\mathcal{o}_x, \mathcal{m}_x)$, and x is a linear function on $\mathcal{m}_x/ \mathcal{m}_x^2$, then the composition x$\circ \phi^*$ will map $\mathcal{m}_y/ \mathcal{m}_y^2$ to $\mathcal{m}_x/ \mathcal{m}_x^2$. Why would it be an element of $\mathcal{T}(Y)_y$, i.e., a map from $\mathcal{m}_y/ \mathcal{m}_y^2$ to itself?
Are there any mistakes in my understanding? Or are there any equivalent definitions that are easier to follow?
Sincere thanks.