Question: "I wonder how can we describe the k-vector space structure on TpX˜ inherited from TpX under this identification?"
Answer: Since any point in a scheme is contained in an open affine subscheme and since the tangent space is defined "locally" we may assume $X$ is affine.
Let $I\in X(k)$ with $X:=Spec(A)$ and $A$ a $k$-algebra ($I:=\mathfrak{m}_p$ in your notation). Given any element $\mu\in Hom_k(I/I^2, R/I)$, there is a direct sum decomposition $R/I^2 \cong k \oplus I/I^2$ and a canonical map
$$ \phi(\mu): R/I^2 \rightarrow k[\epsilon]$$
defined by
$\phi(\mu)(a,b):=a+\mu(b)\epsilon$ where $a\in k$ and $b\in I/I^2$. You get a canonical map
$$\phi_{\mu}: R \rightarrow k[\epsilon]$$
defined by
$$\phi_{\mu}(x):=\phi(\mu)(\overline{x}),$$
where $\overline{x}\in R/I^2$.
It seems that if you choose $\mu , \eta \in Hom_k(I/I^2,R/I)$ you will get the map
$$\phi_{\mu +\eta}(a,b):=a+(\mu+\eta)(b)\epsilon.$$
and for any element $u\in k$
$$\phi_{u\mu}(a,b):=a+(u\mu)(b)\epsilon.$$
This construction seems to induce a $k$-module structure on $ker(X(k[\epsilon])\rightarrow X(k))$ compatible with the one on $(I/I^2)^*$. If $X$ is a group scheme you should be able to determine the kernel of $X(k[\epsilon]) \rightarrow X(k)$: There is a canonical map
$$u:X(k[\epsilon]) \rightarrow X(k)$$
induced by the canonical map
$$R \rightarrow k[\epsilon] \rightarrow k$$
and your point $I\in X(k)$. Hence if you assume $\phi\in X(k[\epsilon])$
with $\phi \in ker(u)$ it follows
$$\phi(a):=\overline{a}1 +\phi_1(a)\epsilon$$
where $\overline{a}\in R/I$ is the class of $a$. Given two maps $\phi, \psi\in ker(u)$ you may "add them" and "multiply with a constant $\alpha$":
$$\rho, \eta:R \rightarrow k[\epsilon]$$
defined by
$$\rho(a):= \overline{a}1+(\phi_1+\psi_1)(a)\epsilon$$
and
$$\eta(a):=\overline{a}1+(\alpha \psi_1(a))\epsilon.$$
This should type of reasoning should induce a $k$-module structure on $\widetilde{T_p(X)}:=ker(u)$ giving an isomorphism of $k$-modules
$$T_p(X) \cong \widetilde{T_p(X)}.$$
Example: $Lie(SL(V))$ Let $V:= k\{e_1,e_2\}$ and $SL(V)$ be the group scheme of linear automorphisms of $V$ with determinant $1$, where $k$ is any commutative unital ring. Let
$$R:=k[x_{11}, x_{12},x_{21},x_{22}]/(det(x_{ij})-1).$$
It follows $SL(V):=Spec(R)$. We want to determine $T_e(SL(V))$ - the tangent space at the identity $e$.
By definition $T_e(SL(V)):=ker(SL(V)(k[\epsilon]) \rightarrow SL(V)(k)):=ker(u)$.
A map $\phi \in ker(u)$ is a map on the form
$$\phi: R \rightarrow k[\epsilon]$$
with $\phi(x_{11}):=1+a_{11}\epsilon, \phi(x_{12}):=a_{12}\epsilon, \phi(x_{21}):=a_{21}\epsilon, \phi(x_{22}):=1+a_{22}\epsilon$ with $a_{ij}\in k$.
The map $\phi$ induce a map $\phi^*:k[x_{ij}]\rightarrow k[\epsilon]$ with the property that $\phi^*(det(x_{ij})-1)=0$, and since the composed map $k[x_{ij}]\rightarrow k$ should correspond to the identity matrix, it follows the induced map $\phi$ must satisfy the above conditions.
Since the map $\phi$ is well defined it follows $\phi(det(x_{ij})-1)=0$ and this holds iff $a_{11}+a_{22}=0$.
Hence $ker(u)$ is the set of matrices $sl(V) :=\{A:=(a_{ij})\in Mat(2,k)$ with $tr(A)=0\}$.
Hence
$$Lie(SL(V))\cong sl(V).$$
The following post considers an intrinsic construction of the structure of a Lie algebra on $Lie(SL(V))$ and the universal envloping algebra $U(Lie(SL(V))$ (you must consider the algebra of first order differential operators and the algebra of distributions):
Distributions of a group scheme as differential operators.
If $G:=Spec(R)$ is a group scheme and $I\subseteq G$ is the ideal of the unit element it follows
$$Dist^1(G,e):= Hom_{R/I}(R/I^2, R/I)$$
has the property that if the base field $k$ is of characteristic zero it follows
$$Dist^n(G,e) \cong U^n(Lie(G))$$
is the $n$'th piece of the canonical filtration of the enveloping algebra $U(Lie(G))$. Hence $Dist^1(G,e)=k\oplus Lie(G)$. Since $Dist^1(G,e) \cong Diff^1_k(R/I^2,k)$ is the module of first order differential operators from $R/I^2$ to $k$, there is a canonical structure of Lie algebra on $k\oplus Lie(G)$ inducing one on $Lie(G)$.
Note: The construction is elementary - you must get used to working with coordinate rings and their maximal and prime ideals. In fact: If $R$ is a Hilbert-Jacobson ring it is enough to work with maximal ideals. Any commutative unital ring that is finitely generated over a field or a Dedekind domain is a Hilbert-Jacobson ring.
