I'm a little confused as to why you'd want to consider the tangent space as a scheme. The scheme structure, if any, would come from the fact that it's a vector space, not because there is some natural scheme structure on it. For example, in topology, one doesn't often consider the cotangent space to be a manifold.
That said, one can ask for a scheme structure on the tangent bundle of a variety (or, more generally, the relative cotangent space of a map $X\to Y$ of schemes). This parallels exactly what one does in the case of topology--one considers the cotangent bundle of a manifold as a manifold.
To define the tangent bundle is a bit involved. For a variety $X/k$ the cotangent bundle is $\mathcal{Spec}(\text{Symm }\Omega_{X/k})$ where $\Omega_{X/k}$ is the cotangent sheaf. This looks a little confusing, but it's because I'm making some identifications. Namely, the tangent sheaf is the dual $\Omega_{X/k}^\vee$, and then the vector bundle associated to that is $\mathcal{Spec}(\text{Symm }(\Omega_{X/k}^\vee)^\vee)$ which is the same thing (in the case $X$ is a variety) as what I wrote above.
I think what you may be asking though is not what the scheme structure of the tangent space is, but what is the vector space structure. For an affine finite type $k$-scheme, and a $k$-rational point (i.e. one of the form $(x-a,y-b,z-c)$) there is a very natural way to describe the space.
Namely, let $X=k[x_1,\ldots,x_n]/(f_1,\ldots,f_r)$ be our affine finite type $k$-scheme and $p=(a_1,\ldots,a_n)=(x_1-a_1,\ldots,x_n-a_n)$ be our point. We obtain a linear map $J_p:k^n\to k^r$ defined by the Jacobian map:
$$J_p=\begin{pmatrix}\frac{\partial f_1}{\partial x_1}(p) & \cdots & \frac{\partial f_1}{\partial x_n}(p)\\ \vdots & \ddots & \vdots\\ \frac{\partial f_r}{\partial x_1}(p) & \cdots & \frac{\partial f_r}{\partial x_n}(p)\end{pmatrix}$$
Then, one can show that $T_{X,p}$ is isomorphic to $\ker J_p$.
This is a good exercise, one I leave to you. I will outline the idea though. First, prove the proposition for $r=0$ (i.e. $X=\mathbb{A}^n$). Then, identify any $X$ (written as above) as the zero set of a map $f:\mathbb{A}^n\to\mathbb{A}^r$. This will allow you to write an "exact sequence" $X\to\mathbb{A}^n\to\mathbb{A}^r$. This will actually be an exact sequence when you move to ideal land. You can then show that $T_{X,p}$ will be the kernel of the induced map $T_{\mathbb{A}^n,p}\to T_{\mathbb{A}_n,f(p)}$ which, when you identify these spaces with $k^n$ and $k^r$ (as you should have in the first step) will just be the map $J_p$.
One can actually identify the tangent space of an affine finite type $k$-scheme $X$ as the kernel of the Jacobian (defined appropriately) for any $p\in X$ where $p$ is a closed point with $k(p)/k$ separable. It fails in the non-separable case: think about $\text{Spec}(\mathbb{F}_p(T^{\frac{1}{p}}))/\mathbb{F}_p$.
Here is an elementary argument. We want to find a perfect pairing of $k$-vector spaces
$$
\mathfrak{m}/(\mathfrak{m}^2 + \mu A) \times \mathrm{Der}_\Lambda(A,k) \to k
$$
or equivalently, a $\Lambda$-linear pairing
$$
\mathfrak{m} \times \mathrm{Der}_\Lambda(A,k) \to k
$$
with trivial right kernel and left kernel equal to $\mathfrak{m}^2 + \mu A$.
The pairing is evaluation of a derivation on an element of $\mathfrak{m}$:
$$
(x, \partial) \mapsto \partial(x) \in k.
$$
Note first that by assumption, the natural map $\Lambda \to A$ induces an isomorphism $\Lambda/\mu \to A/\mathfrak{m}$. In particular, every element of $A$ can be written as $r + x$ where $r$ is in the image of $\Lambda \to A$ and $x \in \mathfrak{m}$.
Suppose now that $\partial$ is a derivation such that $\partial(x) = 0$ for all $x \in \mathfrak{m}$. Then $\partial(r + x) = \partial(r) + \partial(x) = 0$ since $\partial$ is $\Lambda$-linear and so $\partial$ is the $0$ derivation. This proves that the right kernel of the pairing is zero.
Next note that both $\mathfrak{m}$ and $\mu$ act by $0$ on $k$. Therefore, by Leibniz rule and $\Lambda$-linearity, $\partial(\mathfrak{m}^2 + \mu A) = 0$ for all derivations $\partial$. Thus the pairing factors through a pairing
$$
\mathfrak{m}/(\mathfrak{m}^2 + \mu A) \times \mathrm{Der}_\Lambda(A,k) \to k.
$$
To conclude that this pairing is perfect, we need to show that for any nonzero $x$ in $\mathfrak{m}/(\mathfrak{m}^2 + \mu A)$, there exists a derivation $\partial$ with $\partial(x) \neq 0$. Let
$$
\bar{A} = A/(\mathfrak{m}^2 + \mu A)
$$
and denote by $\bar{\mathfrak{m}} = \mathfrak{m}/(\mathfrak{m}^2 + \mu A)$. Then composition with the surjection $A \to \bar{A}$ induces a bijection $\mathrm{Der}_\Lambda(\bar{A},k) \to \mathrm{Der}_\Lambda(A,k)$.
Now $\bar{A}$ is a local Artinian $k$-algebra with maximal ideal $\bar{\mathfrak{m}}$ satisfying that $\bar{\mathfrak{m}}^2 = 0$. Then $\mathrm{Der}_\Lambda(\bar{A},k) = \mathrm{Der}_k(\bar{A},k)$ and we can identify our pairing with the same pairing for $\bar{A}$ as a $k$-algebra, namely
$$
\bar{\mathfrak{m}} \times \mathrm{Der}_k(\bar{A},k) \to k
$$
and now this is the familiar setting where we can check the pairing is perfect by hand, e.g. by picking a splitting $\bar{A} = k \oplus \bar{\mathfrak{m}}$ as vector spaces.
I should note that $\mathfrak{m}/(\mathfrak{m}^2 + \mu A)$ is the relative Zariski cotangent space for the map of schemes $\mathrm{Spec} A \to \mathrm{Spec} \Lambda$ and the above result is true more generally for any morphism of schemes $f : X \to S$. This is proved using more sophisticated ideas in the Stacks Project.
Best Answer
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.