Dot product (of coordinate vectors) is indeed dependent on coordinates. For instance, the vector
$$
e_1
$$
has coordinate vector $v = (1,0,0)$
in the basis $e_1, e_2, e_3$ which has $$v \cdot v = 1,$$
but the same vector, expressed in the basis
$$\frac{1}{2} e_1, e_2, e_3$$
has coordinate vector $$
w =(2, 0, 0),
$$
and $w\cdot w = 4$.
post-comment additions:
To explain the other problem you're having:
When you write "Let
$$ \textbf{v}=R\textbf{e}_r+\Theta\textbf{e}_{\theta}+Z\textbf{e}_z\in \mathbb{R}^3 \ldots,$$
you've already gone off the rails. Because the basis you've written at the top of your post is the basis for the space of "vectors based at the point whose coordinates, in the polar system, are $(r, \theta, z)$." It's not a basis for $\mathbb R^3$, because in the context of $\mathbb R^3$, the values $r$ and $\theta$ aren't even defined!
This'll become obvious once I make things concrete: Let's the the vector $v$ where $R = 2, \Theta = \frac{\pi}{2}, Z = 1$. In the expression for $e_r$ that you plugged in, what's the value of $\theta$? You seem to have plugged in $\Theta$, but why? For the basis to make sense, you need the $(r, \theta, z)$ coordinates of the point at which you're using it as a basis, but you don't have those...so you've used the nearest thing, typographically, as a substitute. There's really no justification for that.
Some Gratuitous Advice
A question for you: When you replaced the lower-case $\theta$ and $r$ with their upper-case versions, did something in the back of your mind say, "Hey, wait a minute...these are actually different!"? And did you then perhaps say "Yeah, but they're the only "r" and "theta" I can see in the formulas I've got, so I guess I have to use them!"? Because that voice in your head was the warning that you were doing something wrong, and needed a deeper understanding before proceeding.
I spend a good deal of time programming, and I find debugging about 10 times as hard as programming. That's pretty much true for math as well, and listening to that little voice is part of the way to avoid debugging (in both contexts).
$\hat{\theta}$ tells you the angle you need to be at from the positive $x$-axis, and $\hat{r}$ tells you how far you need to walk out from the origin. The magnitude is most certainly given by the $\hat{r}$ component. Perhaps your issue is that you're not adding/multiplying vectors in polar form properly. You cannot simply take, for example, $v_1 = 1\hat{r} + \pi\hat{\theta}$, and conclude that $v_1 + v_1 = 2\hat{r} + 2\pi\hat{\theta}$. Notice that the angle has been changed, which shouldn't happen for two vectors that are colinear.
It is better to represent vectors as $re^{i\theta}$. From which in our example we have, $v_1 + v_1 = e^{i\pi} + e^{i\pi} = 2e^{i\pi}$, which has an $\hat{r}$ of 2, and $\hat{\theta}$ of still $\pi$.
Best Answer
Given the Cartesian representation of the force,
$$\vec{F} = F_x\vec{i}+F_y\vec{j}+F_z\vec{k}$$
its cylindrical equivalent is
$$\vec{F} = F_0\cos(\theta-\theta_0)\vec{u}_r-F_0\sin(\theta-\theta_0)\vec{u}_\theta+F_z\vec{k}$$
where $F_0=\sqrt{F_x^2+F_y^2}$ and $\theta_0=\arctan\frac{F_y}{F_x}$.
Note that there is a distinction between the coordinate angle $\theta$ and the angle $\theta_0$ of the force $F$. Their difference produces the circumferential load.