Formula for tangent map, proof check

Consider smooth map $$f:\mathbb{R}^{n}\to \mathbb{R}$$, let $$a \in \mathbb{R}^{n}$$ be any point, $$X \in T_{a}\mathbb{R}^{n}=\mathbb{R}^{n}$$ be tangent vector at point $$a$$.

I have probably proven the following:

Claim: $$df_{a}(X)=\frac{d}{dt}|_{t=0}f(a+Xt)$$, where $$d$$ is the tangent map or derivation of $$f$$.

""Proof"": Let $$h:\mathbb{R}\to \mathbb{R}$$ be some smooth function. Let's evaluate $$df_{a}(X)$$ at $$h$$.

$$df_{a}(X)h = X|_{a}(hf)$$ (this means directional derivative by $$X$$ at point $$a$$. also this point follows from definition)

Now by chain rule, we obtain $$X|_{a}(hf)=\frac{d}{dt}|_{t=f(a)}h \cdot X|_{a}f$$.

I.e. $$df_{a}(X) = X|_{a}f \cdot \frac{d}{dt}|_{t=f(a)}$$.

Because we identify real numbers $$x \in \mathbb{R}$$ with one dimensional derivations $$x \cdot \frac{d}{dt}|_{t=f(a)}$$, we conclude:

$$df_{a}(X) = X|_{a}f$$

Which by the definition equals to $$\frac{d}{dt}|_{t=0}f(a+Xt)$$. Q.E.D.

The whole reasoning seem clumsy and far fetched to me. If not wrong. Is my proof correct? What would you suggest to change/improve? Thanks in advance.

You proof is correct. However, there are various equivalent definitions of the tangent space $$T_a\mathbb{R}^{n}$$. It is isomorphic to $$\mathbb{R}^{n}$$, but you should make precise which definition you use and how the isomorphism $$T_a\mathbb{R}^{n} \to \mathbb{R}^{n}$$ is given.
Obviously you use the description via derivations. In this case $$X \in \mathbb{R}^{n}$$ is identfied with the directional derivation $$D_{a,X}$$ at $$a$$ in direction $$X$$ which you write as $$X \mid_a$$. In the one-dimensional case you say correctly that $$x$$ is identified with $$x \cdot \frac{d}{dt} \mid_{f(a}$$ which is the directional derivation $$D_{f(a),x} = x \mid_{f(a)}$$. Your claim should then precisely be understood as follows:
Under the above identications of tangent spaces, $$df_{a}(X)=\frac{d}{dt}|_{t=0}f(a+Xt) = D_{a,X}(f) = X \mid_a(f)$$.