I'm teaching myself differential geometry, and the author I've been reading opts to work with germs of locally defined functions on a (not necessarily smooth) manifold to describe the tangent space. So, in what follows, $\mathcal{O}^{(k)}_{M,p}$ will denote the set of germs of $C^k$ functions defined in some open neighborhood of $p\in M$, and $\mathcal{S}^{(k)}_{M,p}$ will denote the subset of germs which are stationary at $p$, i.e. where all partial derivatives vanish in any chart. The tangent space $T_p M$ is then defined as the set of linear forms $L:\mathcal{O}^{(k)}_{M,p} \to \mathbb{R}$ which vanish on $\mathcal{S}^{(k)}_{M,p}$, which is isomorphic to the quotient dual $\left( \mathcal{O}^{(k)}_{M,p} / \mathcal{S}^{(k)}_{M,p} \right)^*$.
I'm trying to prove the correspondence between tangent vectors and linear derivations. The partial differentiation operation in $\mathcal{O}^{(k)}_{M,p}$ exhibits a property similar to the Leibniz rule.
$$ \left( \dfrac{\partial}{\partial x_i} \right)_p \left( \mathbf{f}\mathbf{g} \right) = \mathbf{f}(p) \left\lbrace \left( \dfrac{\partial}{\partial x_i} \right)_p \mathbf{g} \right\rbrace + \left\lbrace \left( \dfrac{\partial}{\partial x_i} \right)_p \mathbf{f} \right\rbrace \mathbf{g}(p)$$
Strictly speaking, a derivation should be of the form $D:R\to M$ where $R$ is a ring and $M$ is an $R$-bimodule (possibly $R$ itself). In the above, the pesky evaluation at the point $p$ serves to get in the way. We can consider the partial derivative map as a binary operator
$$ \left( \dfrac{\partial}{\partial x_i} \right) : \mathcal{O}^{(k)}_{M,p} \times U \to \mathbb{R} : (\mathbf{f},q) \mapsto \left( \dfrac{\partial}{\partial x_i} \right)_q \mathbf{f} $$
where $U$ is any open neighborhood of $p$. Then the Leibniz rule can take its proper form.
$$ \left( \dfrac{\partial}{\partial x_i} \right) \left( \mathbf{f}\mathbf{g} \right) = \mathbf{f} \left\lbrace \left( \dfrac{\partial}{\partial x_i} \right) \mathbf{g} \right\rbrace + \left\lbrace \left( \dfrac{\partial}{\partial x_i} \right) \mathbf{f} \right\rbrace \mathbf{g} $$
My question is this: how can we show that every tangent vector $u\in T_p M$ corresponds bijectively to a derivation $D: \mathcal{O}^{(k)}_{M,p} \to \mathcal{O}^{(k)}_{M,p}$? How might we extend this to $\mathcal{O}^{(k)}_{M,U}$, the germs of functions defined on some open subset, or eventually to $C^{(k)}(M,\mathbb{R})$, the global functions on $M$?