When calculating the velocity in polar coordinates, it will be the following
$$ v=\dot{r}e_{r}+r\dot{\theta}e_{\theta}.$$
But, for acceleration, we take the derivative of the basis vector as well
$$a=\ddot{r}e_{r}+\dot{r}\dot{e}_{r}+\dot{r}\dot{\theta}e_{\theta}+r\ddot{\theta}e_{\theta}+r\dot{\theta}\dot{e}_\theta.$$
And I was wondering why do we take the derivative of the basis during acceleration and not during velocity as well.
As Ghoster pointed out in the comments, I believe your question is based on a misunderstanding, which has to do with confusing between vector components and coordinates.
The confusion arises here in relation to the position vector, $\vec{r}$. When expressed in the Cartesian basis, there is indeed a one-to-one correspondence between the point in space we want to express via the position vector, and its vector components. So that for $2D$, the position vector representing the point $(x,y)$ in the Cartesian plane is uniquely given by:
$$\vec{r} = x\hat{e}_x + y\hat{e}_y, \tag{1} $$
then, taking the time derivative to find the velocity we find:
$$\dot{\vec{r}} = \dot{x}\hat{e}_x + \dot{y}\hat{e}_y \tag{2} $$
where we note that, because the basis vectors are everywhere constant, $\dot{\hat{e}}_x = \dot{\hat{e}}_y = 0$.
But, the uniqueness of $(1)$, which relates position vector components to points in space does not hold for other basis sets in general!
To see why, note that for example in the $2D$ polar coordinates you've mentioned, the basis vectors are in fact vector fields, they depend on the angle $\theta$. We see this if we express them in terms of the Cartesian basis:
$$ \hat{e}_r(\theta) = \cos\theta \hat{e}_x + \sin\theta \hat{e}_y $$ $$ \hat{e}_{\theta}(\theta) = -\sin\theta \hat{e}_x + \cos\theta \hat{e}_y $$
This means, among other things, that there isn't a one-to-one correspondence between the components of a position vector and the actual position it gives. For example, as you may verify, we can represent the Cartesian position $(1,1)$ via either:
$$ \vec{r} = \sqrt{2}\ \hat{e}_r\big(\small\pi / 4\normalsize\big), $$ or $$ \vec{r} = \sqrt{2}\ \hat{e}_{\theta}\big(-\small\pi / 4\normalsize\big).$$
Now, we have a useful convention in polar coordinates, of always writing the position vector of an arbitrary point $(x,y)$ as:
$$ \vec{r} = r\hat{e}_r, \tag{3} $$
without explicitly specifying the angle $\theta$ that $\hat{e}_r$ depends on, simply because we know that for each position, there exists a suitable $\theta$ that gives $\hat{e}_r$ its right direction. We also want to write position vectors in this basis as proportional to $\hat{e}_r$, which always points radially "outwards" from the origin towards the required point of interest, because then when we then differentiate them with respect to time, the resulting velocity vector's $\dot{\vec{r}}$ radial component that's proportional to $\hat{e}_r$ will be a measure of radial velocity, and its tangential component proportional to $\hat{e}_\theta$ will be a measure of tangential velocity, where "radial" and "tangential" here are relative to the coordinate origin.
So, writing a position vector as in $(3)$ isn't just an arbitrary convention, but a rather useful one to stick to.
Indeed, when we differentiate $(3)$ with respect to time we have:
$$ \dot{\vec{r}} = \dot{r}\hat{e}_r + r\dot{\hat{e}}_r = \dot{r}\hat{e}_r + r\dot{\theta}\hat{e}_{\theta}, \tag{4} $$
as you've correctly written, but the upshot here is that the differentiation of $(3)$ which gives us $(4)$ is not a consequence of some general rule saying "two dimensions, hence two basis vectors always appear", which I think is the source of your issue here. It may look at first sight that in getting $(4)$ "we left the basis vectors alone", because you may think that a general position vector in polar coordinates is of the form $C_r\hat{e}_r + C_\theta \hat{e}_{\theta} $, in direct analogy to $(1)$. As I hope I managed to explain in this answer, this isn't the case.
We do take the derivatives of basis vectors when calculating velocity. Suppose the cartesian basis vector $\hat{x}$ an $\hat{y}$ is not moving with $$\hat{e}_r=\cos(\theta)\hat{x}+\sin(\theta)\hat{y}$$ $$\hat{e}_\theta=-\sin(\theta)\hat{x}+\cos(\theta)\hat{y}$$ then it can be shown by differentiating both sides to get $$\frac{d \hat{e}_r}{dt}=\dot{\theta}\hat{e}_\theta$$ $$\frac{d \hat{e}_\theta}{dt}=-\dot{\theta}\hat{e}_r$$ It's clear the polar basis vectors are moving. Since $$\mathbf{r}=r\hat{e}_r$$ We can get the velocity in polar coordinate as $$\dot{\mathbf{r}}=\dot{r}\hat{e_r}+r\dot{\theta}\hat{e_\theta}$$ as expected.
We do differentiate unit vectors when calculating velocity. Your first line has $$v=\dot{r}e_{r}+r\dot{\theta}e_{\theta}$$ but if you look at the steps in between
$$\mathbf{v} = \frac{d}{dt} (r \mathbf{e}_r)$$
$$\rightarrow\mathbf{v} = \dot{r} \mathbf{e}_r + r \dot{\mathbf{e}}_r $$
$$\mathbf{v} = \dot{r} \mathbf{e}_r + r \dot{\theta} \mathbf{e}_\theta$$
where $$\dot e_r=\frac{d{\mathbf{e}}_r}{dt} =\dfrac { d\mathbf e_r} { d\theta} \dfrac {d\theta} { dt}= \dot{\theta} \mathbf{e}_\theta$$
Also, after the first derivative, $\bf e_\theta$ is now explicit in $\bf v$ and so when calculating acceleration, we must differentiate $\bf e_\theta$ as well as ${\mathbf{e}}_r$:
$$\mathbf{a} = \frac{d}{dt} \left( \dot{r} \mathbf{e}_r + r \dot{\theta} \mathbf{e}_\theta \right)$$
so that
$$ \mathbf{a} = \ddot{r} \mathbf{e}_r + \dot{r} \dot{\mathbf{e}}_r + \dot{r} \dot{\theta} \mathbf{e}_\theta + r \ddot{\theta} \mathbf{e}_\theta + r \dot{\theta} \dot{\mathbf{e}}_\theta$$
The basis vectors ($\mathbf{e}_r$ and $\mathbf{e}_\theta$) indirectly depend on $t$ because their directions vary with the angle $\theta$. You need to consider this when applying the product rule for calculating the time derivatives of $\dot{r}\mathbf{e}_r$ and $r\dot{\theta}\mathbf{e}_\theta $.
$$\begin{align} \mathbf{a} &= \frac{d\mathbf{v}}{dt} \\ &= \frac{d}{dt}(\dot{r}\mathbf{e}_{r}+r\dot{\theta}\mathbf{e}_{\theta}) \\ &= \frac{d}{dt}(\dot{r}\mathbf{e}_{r})+\frac{d}{dt}(r\dot{\theta}\mathbf{e}_{\theta}) \\ &= (\ddot{r}\mathbf{e}_r+\dot{r}\dot{\mathbf{e}}_r) +(\dot{r}\dot{\theta}\mathbf{e}_\theta+r\ddot{\theta}\mathbf{e}_\theta+r\dot{\theta}\dot{\mathbf{e}}_\theta) \end{align}$$
