Silly question about basis vectors as partial derivatives It's good to construct a notion of basis of a particular coordinate system as (for instance, in polar coordinate system):
$$\displaystyle \frac{\partial\vec{R}}{\partial r}= \frac{\vec{R_{h}}(r+h,\theta) - \vec{R}(r,\theta)}{h}:= \vec{e}_{r}$$ 
$$\displaystyle \frac{\partial\vec{R}}{\partial \theta}= \frac{\vec{R_{h}}(r,\theta+h) - \vec{R}(r,\theta)}{h}:= \vec{e}_{\theta}$$
Because now we have a quite general notion of basis of a coordinate system. But what is this "general notion of basis"? Well, in that question lies my doubt.
When is defined (in a elementary calculus course for example) the notion of partial derivatives,they are defined upon scalar functions like $f(r,\theta)$ i.e. Scalar Fields. 
But in order to view basis vectors as partial derivatives, we defined up above as partial derivatives of a Vector $\vec{R}$. Furthermore, the books said that actually these partials ($\displaystyle \frac{\partial\vec{R}}{\partial \theta}$, $\displaystyle \frac{\partial\vec{R}}{\partial r}$) are indeed tangents (vectors) to coordinate curves.
So it seems that the whole construct deals with, Position vectors $\vec{R}$, a particular curve (parametrization) $\gamma (\lambda)$ and the composition:
$$\vec{R} \circ \gamma (\lambda) = \vec{R} [\gamma (\lambda)] = \vec{R}[r(\lambda),\theta(\lambda)]$$
And then, the "true" meaning of partial derivatives as basis vectors lies in this kind of derivation:
$$\displaystyle \frac{\partial}{\partial r}[\vec{R} \circ \gamma (\lambda)]:= \vec{e}_{r}$$ 
$$\displaystyle \frac{\partial}{\partial \theta} [\vec{R} \circ \gamma (\lambda)] := \vec{e}_{\theta}$$
Is it right what I've said?
 A: You are correct until the last two definition $$\frac{\partial}{\partial r}[\vec{R} \circ \gamma (\lambda)]:= \vec{{e}_{r}}$$ 
$$\frac{\partial}{\partial \theta} [\vec{R} \circ \gamma (\lambda)] := \vec{{e}_{\theta}}$$
because the expression $\vec{R} \circ \gamma (\lambda)$ depends on solely $\lambda$. I don't think you would want to take partial derivative with respect to $r$ or $\theta$.
Instead, you have two curves $\gamma_r,\gamma_\theta$. The first curve will be used to define $\vec{{e}_{r}}$ and the other will be used to define $\vec{{e}_{\theta}}$. The curves are defined as $$\gamma_r:(0,\infty)\to(0,\infty)\times\Bbb R,\ \ \ \gamma_r(\lambda)=(\lambda,\theta_0)$$ where $\theta_0$ is a constant, and $$\gamma_\theta:\Bbb R\to(0,\infty)\times\Bbb R, \ \ \ \gamma_\theta(\lambda)=(r_0,\lambda)$$ where $r_0$ is a constant.
Then $$\frac{d}{d\lambda}[\vec{R} \circ \gamma_r (\lambda)]= \vec{{e}_{r}}$$ $$\frac{d}{d\lambda}[\vec{R} \circ \gamma_\theta (\lambda)]= \vec{{e}_{\theta}}$$
On a closer inspection, this is no different from the first definition you stated.
