I'm studying introductory vector calculus and need to confirm/clarify my concepts. The definition of the derivative of a vector (for example in $\mathbb{R}^2$) if the unit vectors are constant throughout the 2D space is in terms of its components: if we have $r(t)=(x(t), y(t))$ in the standard Cartesian basis, then
$$\frac{dr}{dt}=\frac{dx}{dt}e_x+\frac{dy}{dt}e_y$$
Now if we move to polar coordinates $\rho, \phi$, then the unit basis vectors $e_{\rho},e_{\phi}$ will change direction depending on the location in 2D space. To define the derivative in this case, the book that I'm studying gives the following quick method: we see that $r=\rho e_{\rho}$ (where $\rho$ is the distance of the vector's endpoint from the origin), which means
$$\frac{dr}{dt}=\frac{d\rho}{dt}e_{\rho}+\rho\frac{de_{\rho}}{dt}$$
So far, so good: $\frac{d\rho}{dt}$ can be calculated since we can express $\rho$ in terms of $x(t)$ and $y(t)$, and differentiate that expression w.r.t. $t$. In this specific case, we can also express $e_{\rho}=(\cos\phi)e_x + (\sin\phi)e_y$. It turns out that $$\frac{de_{\rho}}{dt}=\frac{d\phi}{dt}e_{\phi}$$ because of the specific way $e_{\rho}$ and $e_{\phi}$ are defined in terms of $e_x$ and $e_y$.
Expressing the same vector $r$ in a general curvilinear coordinate system $u,v$,
To even start differentiating $r$, we need to find the components of $r$ in the new system. I'm assuming the way to identify $r$ is to identify it as the intersection of two coordinate curves $u=c_1$ and $v=c_2$ - in this case, $u=5$ and $v=4$. Is my understanding correct? Is this the way to identify the components of a vector in a curvilinear system?
So if we have some differentiable functions $f,g$ such that $u=f(x,y)$ and $v=g(x,y)$ and $r=ue_u+ve_v$, then $$\frac{dr}{dt}=\frac{du}{dt}e_u+u\frac{de_u}{dt}+\frac{dv}{dt}e_v+v\frac{de_v}{dt}$$
$\frac{du}{dt}$ can be identified as $\frac{df(x(t),y(t))}{dt}$ and can be evaluated. How does one, in general, express basis vectors $e_u$ and $e_v$ in terms of $e_x$ and $e_y$? And even if we do manage to define curvilinear basis vectors in terms of $e_x,e_y$, it's not necessary that we'll get a nice expression for $\frac{de_u}{dt}$ and $\frac{de_v}{dt}$ in terms of $e_u$ and $e_v$. How do we get the curvilinear components of $\frac{dr}{dt}$ in that case?