The definition for the differential of a morphism 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.
 A: The definition I prefer is the following: a tangent vector $v$ to a point $p : k[X] \to k$ on $X$ is a morphism $k[X] \to k[\epsilon]/\epsilon^2$ such that quotienting by $\epsilon$ gives the point $p$ back. (It sends a function $f \in k[X]$ to $f(p) + \epsilon f_v(p)$ where $f_v$ denotes the directional derivative in the direction $v$.) The collection of all tangent vectors at $p$ is the tangent space, and composing with a morphism $k[Y] \to k[X]$ turns a tangent vector on $X$ into a tangent vector on $Y$ in a completely transparent way. (This definition, as it turns out, also works for smooth manifolds.) 
A: If $\lambda$ is an element in $T_x=(m_x/m_x^2)^\ast$ the composition $\lambda\circ\phi^*$ is a linear functional on $m_{\phi(x)}/m_{\phi(x)}^2$, i.e. an element of $T_y$ where $y=\phi(x)$.
The differential $d\phi_x$ maps $\lambda$ to $\lambda\circ\phi^*$.
This is nothing but the fact that passing to the dual switches the arrows, so to speak. If you have a linear map $f:V\rightarrow W$, the dual map goes $f^*:W^*\rightarrow V^*$.
