As in the last time the first issue came up, you have misread Hartshorne's definition and Hartshorne failed to be as specific as he ought to have been. Hartshorne's text actually says "$k(x)=k$," which is intended to mean that the map is the canonical map $k\to \mathcal{O}_{X,x}/\mathfrak{m}_x$ you write. (A less confusing way to write this may be to specify that a $k$-rational point for a scheme over $k$ is a map $\operatorname{Spec} k\to X$ so that the composition with the structure map $X\to \operatorname{Spec} k$ is the identity.)
Next, Wikipedia actually gives two definitions: the first which depends on an embedding and a choice of algebraic closure, and then a second which is the same as the Hartshorne definition. The first is (at least to me) a little poorly explained: it leaves open the possibility that for a variety over $k$ you could ask about $l$ points for $l$ a subfield of $k$, which is not a thing. Such $l$ points aren't intrinsically defined (for instance, the point $(x-1)\in \Bbb A^1_\Bbb C$ can be moved to $(x-i\pi)\in \Bbb A^1_\Bbb C$ by several obvious automorphisms) in contrast to asking about points corresponding to extension fields.
Your question about looking at the composite map $X\to \operatorname{Spec} k\to \operatorname{Spec} k'$ is reasonable, but it fails to give you any interesting answer. No point of a $k$-variety will be a $k'$-point unless $k'=k$ (and even then you may not have any). The fact that $X$ is a $k$-variety says that the residue field of every point of $X$ is some (possibly trivial) extension of $k$, and unless $k=k'$, none of these can be $k'$. For instance, in your final example, there are no points on the right hand side of your equation in your second-to-last line.
To understand why these two definitions coincide, it's enough to work affine-locally. Then we have both an embedding $X\to \Bbb A^n_k$ and a map $\operatorname{Spec} F\to X$ picking out the $F$-point. Writing down the map on rings corresponding to the composition $\operatorname{Spec} F\to \Bbb A^n_k$, this is a map $k[x_1,\cdots,x_n]\to F$, and we can think of the coordinates of our $F$-point as the images of $x_1,\cdots,x_n$ in $F$. As the
The functor-of-points perspective is very useful here in giving a clean answer to your question. When $X$ is a scheme and $R$ is a ring, write $X(R)=\mathrm{Hom}_{\mathbf{Sch}}(\mathrm{Spec}(R), X)$. We have a morphism of sets $X(k[\epsilon]/\epsilon^2)\to X(k)$ induced by the morphism of rings $k[\epsilon]/\epsilon^2\to k$ induced by mapping $\epsilon\mapsto 0$. The set-theoretic preimage over some $p\in X(k)$ is your $\widetilde{T_pX}$. We will equip this set with the structure of a $k$-vector space.
First we will define addition on $\widetilde{T_pX}$. In preparation, we want to define a map
$$
+\colon X(k[\epsilon]/\epsilon^2)\times_{X(k)}X(k[\epsilon]/\epsilon^2)\to X(k[\epsilon]/\epsilon^2).
$$
Now,
$$
X(k[\epsilon]/\epsilon^2)\times_{X(k)}X(k[\epsilon]/\epsilon^2)
\simeq X(k[\epsilon]/\epsilon^2\otimes_kk[\epsilon]/\epsilon^2)\simeq X(k[\epsilon_1,\epsilon_2]/(\epsilon_1^2,\epsilon_2^2)).
$$
Thus to define the map $+$, it's enough to define a map on rings. Take the map on rings given by $\epsilon_1\mapsto\epsilon$ and $\epsilon_2\mapsto\epsilon$. The map $+$ gives a morphism over $X(k)$ that is vector addition in the tangent space.
Scalar multiplication works similarly, along the lines you had thought: Given an element $a\in k$, we get a morphism of $k$-algebras $k[\epsilon]/\epsilon^2\to k[\epsilon]/\epsilon^2$ given by $\epsilon\mapsto a\epsilon$. Applying the functor $X(-)$ gives a morphism
$$
X(k[\epsilon]/\epsilon^2)\to X(k[\epsilon]/\epsilon^2)
$$
over $X(k)$ (that is, the triangular diagram I would like to be able to draw here commutes).
Edit:
I apologize as these sections I think are worse than the above. The main point is that if $A$ is a $k$-algebra and you fix a morphism $\phi\colon A\to k$ of $k$-algebras, then a morphism $A\to k[\epsilon]/\epsilon^2$ of the form $a\mapsto\phi(a)+\psi(a)\epsilon$ is just the data of a $k$-derivation $\psi$ of $A$.
Below we'll dispense with what happens at least for linear algebraic groups over the field $k$. (Note that affine implies linear and this comparison makes sense only if $G$ is taken over $k$.)
$\widetilde{T_pX}$ and $T_pX$
This is essentially what the other answer discussed. On the level of sets, the isomorphism $\widetilde{T_pX}\to T_pX$ is given as follows. Let $f\in\widetilde{T_pX}$. Then $f$ is a morphism
$$
f\colon\mathrm{Spec}k[\epsilon]/\epsilon^2\to X
$$
such that the only closed point $(0)$ of the source is sent to $p\in X$. We get a morphism of local rings
$$
f^{\#}\colon \mathcal{O}_{X,p}\to k[\epsilon]/\epsilon^2
$$
that must send $\mathfrak{m}_p\to \epsilon k[\epsilon]/\epsilon^2\simeq k$.
This is clearly the same as map $\mathfrak{m}_p/\mathfrak{m}_p^2\to k$. Moreovoer, a map $\mathfrak{m}_p/\mathfrak{m}_p^2\to k$ is the same as a $k$-derivation of $\mathcal{O}_{X,p}$, and addition of linear functionals correspdonds to addition of derivations under this bijection. Explicitly, given a linear functional $\lambda$, one defines $d_\lambda(x)=0$ if $x$ is a unit, and $d_\lambda(x)=\lambda(x)$ if $x\in\mathfrak{m}_p$, and this is a $k$-derivation.
As for the map $+$, if we choose an affine open $\mathrm{Spec}A$ of $X$, containing $p$, we see that the sum of two elements of $\widetilde{T_pX}$
is a ring morphism
$$
A\to k[\epsilon_1,\epsilon_2]/(\epsilon_1^2,\epsilon_2^2)\to k[\epsilon]/\epsilon^2.
$$
By definition of $+$, the derivation this ring morphism encodes (the function giving the coefficient of $\epsilon$) is the sum of the derivations we started with. This is the statement that the the isomorphis of sets above is an isomorphism of abelian groups. Scalar mutliplication is similar.
The case of linear algebraic groups
Let $G$ be a linear (equivalently, affine) algebraic $k$-group, and define
$$
\mathrm{Lie}_1(G):=\mathrm{ker}(G(k[\epsilon]/\epsilon^2)\to G(k)).
$$
Thus $\mathrm{Lie}_1(G)$ is a group, although a priori it is not obviously an abelian group. As just the kernel of a group homomorphism, it is not a priori a vector space over $k$. We'll try to show that it's isomorphic as a group to your $T_pX$, and then that it has a natural $k$-action that makes this isomorphism one of $k$-vector spaces. Then we'll conclude from the section below. This material is totally standard and in Borel, Springer, Milne, various books of Conrad and co-authors etc, but nowhere did I find the precise statement that I need below (at least, not yet).
Recall that your $T_pX$ is isomorphic as a $k$-vector space to the space of $k$-derivations of $\mathcal{O}_{X,p}$. Indeed, given a morphism $\varepsilon\colon A\to k$ of $k$-algebras, let $\mathfrak{m}=\ker\varepsilon$. Then $\mathrm{Der}_k(A)\to(\mathfrak{m}/\mathfrak{m}^2)^*$ sending $\delta\mapsto\delta|_{\mathfrak{m}}$ is a linear isomorphism.
Define
$$\mathrm{Lie}(G):=T_eG.$$
We will define an isomorphism $\mathrm{Lie}_1(G)\to\mathrm{Lie}(G)$ of groups. Indeed, a morphism of rings $A\to k[\epsilon]/\epsilon^2$ is given by an morphism of $A$-algebras $A\to k$ and a $k$-derivation of $A$. For such a morphism to belong to $\mathrm{Lie}_1(G)$ means that the morphism of $k$-algebras is always the counit $\varepsilon\colon A\to k$ of the Hopf algebra $A$. Thus if $g_1$, $g_2$ are in the kernel, their product is $$
g_1g_2\colon A\to A\otimes_kA\to k[\epsilon]/\epsilon^2
$$
and is also in the kernel. Thus the corresponding morphism $A\to k$ is again the counit, and the only question is what the derivation is. Of course, it just has to be the sum of the derivations for the $g_i$, but I'm dense enough to not extract this immediately from the slightly different treatments I listed above.
Best Answer
Source: Bryden R. Cais' online solution.
Denote $A=k[\epsilon]/\epsilon^2$ the regular local ring. We can see that $A$ has only one prime ideal, namely $m=\epsilon A$ the maximal ideal. Each element of $m$ is in the form of $\lambda\epsilon,\lambda\in k$. Use the isomorphism $m/m^2\to k,\lambda\epsilon\mapsto \lambda$ to identify the two. Let $\eta\in \mathrm{Spec} A$ be the closed point.
1.Given $f:\mathrm{Spec} A\to X$ a morphism, we set $x=f(\eta)$, then local homomorphism $f_{\eta}^{\#}:O_x\to A_m=A$ induces a $k-$morphism of residues fields $k(x)\to k$, hence an isomorphism. So, $x$ is a rational point. And $f_{\eta}^{\#}$ induces $m_x/m_x^2\to m/m^2=k$, which is an element of $T_x$.
2.Conversely, let $j:k\to O_x$ and $\phi:O_x\to O_x/m_x$ be the natural maps. As $x\in X$ is rational, $\phi j$ is an isomorphism with the inverse $\Phi:O_x/m_x\to k$. Now given $\psi:m_x/m_x^2\to k$ an element of $T_x$, we define a ring morphism $O_x\to A,h\mapsto \Phi\phi(h)+\psi(h-j\Phi\phi(a))\epsilon$, and this gives us $\mathrm{Spec} A\to \mathrm{Spec} O_x$. There's a natural morphism $\mathrm{Spec} O_x\to X$, thus by composition we get $\mathrm{Spec} A\to X$.