This is to follow up on Frederik Meyer's comment:
Let $X$ be variety over a field $k$, and let $x\in X(k)$. Then, there is a canonical bijection between morphisms (over $k$) $\text{Spec}(k[\varepsilon])\to X$, whose underlying point hits $x$, and $T_x X$.
Here I just use $k[\varepsilon]$ for the dual numbers. And, I use the notation $T_x X$ for the tangent space $(\mathfrak{m}_x/\mathfrak{m}_x^2)^\vee$.
To see this, note that obviously any such map $\text{Spec}(k[\varepsilon])\to X$ must factor uniquely as $\text{Spec}(k[\varepsilon])\to\text{Spec}(\mathcal{O}_{X,x})\to X$. But, maps $\text{Spec}(k[\varepsilon])\to\text{Spec}(\mathcal{O}_{X,x})$ correspond bijectively to $k$-algebra maps $\mathcal{O}_{X,x}\to k[\varepsilon]$.
But, such $k$-algebra maps correspond to $k$-derivations $\mathcal{O}_{X,x}$ 'at $x$', which are maps $\phi:\mathcal{O}_{X,x}\to k$ such that
$$\phi(fg)=\overline{f}\phi(g)+\overline{g}\phi(f)$$
where the bar denotes passage to the quotient $\mathcal{O}_{X,x}\to k$ (why?). But, I claim that such derivations correspond bijectively with $T_{X,x}$.
In particular, suppose that $\phi:\mathcal{O}_{X,x}\to k$ was such a derivation. Note that by restriction we obtain a map $\phi:\mathfrak{m}_x\to k$. But, if $f,g\in\mathfrak{m}_x$, then
$$\phi(fg)=\overline{f}\phi(g)+\overline{g}\phi(f)=0$$
since $\overline{f}=\overline{g}=0$. So, $\phi$ kills $\mathfrak{m}_x^2$, and so we get a well-defined map $\phi:\mathfrak{m}_x/\mathfrak{m}_x^2\to k$, which, since this is $k$-linear, is an element of $(\mathfrak{m}_x/\mathfrak{m}_x^2)^\vee$.
Conversely, suppose that I start with an element $\phi\in (\mathfrak{m}_x/\mathfrak{m}_x^2)^\vee$. Then, how do we extend it to a derivation of $\mathcal{O}_{X,x}$ at $x$? Well, let $f\in\mathcal{O}_{X,x}$ be arbitrary. Suppose that $\overline{f}=\alpha\in k$, then $\alpha\in\mathcal{O}_{X,x}$ and $f-\alpha\in\mathfrak{m}_x$. So, define $\phi(f):=\phi(f-\alpha+\mathfrak{m}_x^2)$. I leave it to you to check that this is a derivation of $\mathcal{O}_{X,x}$ at $x$, and that it's the inverse of the map from the last paragraph. Thus, the gray box is proven.
For you, this is helpful since any map $\text{Spec}(\mathbb{C}[\varepsilon])\to X$ (for any variety $X/\mathbb{C}$) over $\mathbb{C}$ must hit a closed point. But, since $\mathbb{C}$ is algebraically closed, closed points are the same as $\mathbb{C}$-points. Thus, the above shows that, in some sense
$$\text{Hom}_{\text{Spec}(\mathbb{C})}(\text{Spec}(\mathbb{C}[\varepsilon]),X)=\bigcup_{x\in X^\circ}T_{X,x}$$
where $X^\circ$ denotes the closed points of $X$(=$\mathbb{C}$-points). So, if you squint hard enough, the set of all such maps is like the tangent bundle.
The intuition for this, by the way is as follows. A mapping $\text{Spec}(\mathbb{C}[\varepsilon])\to X$ is essentially picking out a closed point, and then putting a little 'fuzz' (non-reduced data) which corresponds to put an infinitesimal direction, or tangent vector, on that point.
Remark: Before you ask if this is what you're supposed to put on an exam, I have no idea. That's between you and you're professor, the above is just one way to think about it. Good luck. :)
