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I am reading from Topology and Geometry by Bredon.

For each smooth map $\gamma: \mathbb{R} \to X, \gamma(0) = p$ associate to it a 'derivation' $D_{\gamma}:\mathcal{O}_{X,p} \to \mathbb{R}$ given by $f \mapsto \frac{d}{dt}f(\gamma(t)) |_{t=0}$. The collection of derivations given by smooth curves is called the tangent space at $p$. I wish to show that this is a vector space without using local coordinates.

The collection of all derivations on $\mathcal{O}_{X,p}$ is clearly a vector space where $(c D)(f)= c \ D(f)$ and $(D_1 + D_2)(f) = D_1 (f) + D_2 (f)$. However it is unclear how the tangent vector derivations are closed under these operations. I should mention that Bredon notes that all derivations are induced from smooth curves 'in the $C^\infty$ case', but I don't want to use this.

Scalar multiplication is given by $c D_{\gamma(t)} = D_{\gamma{(ct)}}$, but the obvious try for vector addition of tangents assumes that some neighborhood of $p$ has a vector space structure, but this structure I think depends on our choice of local isomorphism with $\mathbb{R}^n$. I am trying to formulate these definitions without mentioning or choosing an isomorphism, mirroring the development in Algebraic Geometry.

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    I don't see how you can do this without using the vector-space structure of $\Bbb{R}$. What you have is that $\gamma_1,\gamma_2 : \Bbb{R} \to X$ are smooth curves passing through $p$ at $t=0$; now, if $v_1,v_2$ are the tangent vectors to the respective paths at $t=0$, that is, $$ v_1 = \frac{d\gamma_1}{dt} \bigg|_{t=0},\quad v_2 = \frac{d\gamma_2}{dt} \bigg|_{t=0}, $$ then you want a curve $\gamma_3 : \Bbb{R} \to X$ passing through $p$ at $t = 0$ such that $$ \frac{d\gamma_3}{dt}\bigg|_{t=0} = v_1 + v_2. $$ – 2018-12-06
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    Clearly, you want to say $\gamma_3 = \gamma_1 + \gamma_2$, when you're working in a coordinate neighbourhood of $p$, so that $p$ maps to the origin. Without viewing it like this, I don't see how you can ensure that there is such a curve $\gamma_3$. It is necessary to appeal to the vector space structure of $\Bbb{R}$ in my opinion. – 2018-12-06

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