Let's look at the general acceleration vector in polar coordinates$^*$:
$$\mathbf a=\left(\ddot r-r\dot\theta^2\right)\hat r+\left(r\ddot\theta+2\dot r\dot\theta\right)\hat\theta$$
If we want our object to remain on the same circle, which I'm assuming this is what you are interested in, we must have $\dot r=0$ and $\ddot r=0$. This means our acceleration must have the form:
$$\mathbf a=-r\dot\theta^2\hat r+r\ddot\theta\hat\theta$$
Since, by Newton's laws, the acceleration vector is proportional to the force via the mass $m$ of our particle, we see that we need a radial force magnitude$^{**}$ of
$$F_r=-mr\dot\theta^2$$
and a tangential force magnitude of
$$F_\theta=mr\ddot\theta$$
Because we are confined to move around a single circle we can determine the role of each force. The radial force must only be responsible for changing the direction of the velocity, since if it could effect the speed then this means the object would have to change its $r$ coordinate, hence knocking us off the circle. Similarly, the tangential force must only be responsible for changing the speed of the particle as it moves around the circle, since if it could effect the direction of the velocity then it would do so by knocking us off the circle.
You can probably tell by now that if we want to stay on the circle, these forces are required to be "linked", in a sense. Indeed, if you take the time derivative of $F_r$ and use the requirement that $\dot r=0$, you will find that
$$\frac{\text dF_r}{\text d t}=-2\dot\theta F_\theta$$
showing that the presence of a tangential force requires a change in the magnitude of the radial force in order for the particle to remain on the circle (or, on the flip side, a change in the magnitude of the radial force must be accompanied by a tangential force).
What if this condition is not met? Well, looking at the derivative of $F_r$ without the condition that $\dot r=0$, we must have
$$-m\dot r\dot\theta^2\neq 0$$
This means that $\dot r\neq0$, which means we are no longer engaging in circular motion. Therefore, we need these two force components to be linked in this way to keep the motion circular.
A subtle point remains to be cleared up (as realized in comments to other answers). This does not necessarily mean that these forces are physically linked in general. This answer assumes we have an object undergoing non-uniform circular motion, and then investigates what must be true about the forces acting on the object. However, there could be situations where the radial and tangential force are not physically linked, at which point to achieve non-uniform circular motion you would have to make these forces act in such a way so that non-uniform circular motion is achieved.
$^*$ dots represent a rate of change with respect to time if you are not familiar with calculus. for example, $\dot r$ is the rate of change of the variable $r$ with respect to time. The derivation of this equation can be found here.
$^{**}$ Note that this is what you usually encounter in your introductory physics classes as $F_{r}=mv^2/r$, since for motion along a circle of radius $r$, $\dot\theta=v/r$. The negative sign in this answer is to keep track of the direction of increasing/decreasing $r$, but if you are working a question where you only care about the magnitude of the radial force then this is irrelevant, hence why you usually don't see the negative sign.
Best Answer
Yes. You are treating the velocity as though it were constant over a finite time and it is not. The velocity varies over time and can only be treated as constant over an infinitesimal time. But infinitesimals do not work the way you wrote.
Since $|\vec v|=\sqrt{\vec v^2}=\sqrt{\vec v \cdot \vec v}$ we can write: $$d|\vec v|=|\vec v + d\vec v|-|\vec v|$$ $$ = \sqrt{\left(\vec v + d \vec v \right)^2} - \sqrt{ \vec v^2}$$ the products of infinitesimals drop out and we have $$ = \sqrt{\vec v^2 + 2 \vec v \cdot d\vec v} - \sqrt{\vec v^2}$$ we can series expand the first radical to get $$ = \sqrt{\vec v^2} + \frac{\vec v \cdot d\vec v}{\sqrt{\vec v^2}} - \sqrt{\vec v^2}$$ $$d|\vec v|=\hat v \cdot d\vec v$$ So, $d|\vec v|$ can be zero if $d\vec v$ is perpendicular to $\vec v$.
Note, in all of the above derivations the vector $\vec v+ d\vec v$ is obtained from the standard Euclidean vector addition. $\vec v$ is infinitesimally different from $\vec v + d\vec v$ in the usual way.
Edit: for more information on how infinitesimals, differentials and so forth work see: https://people.math.wisc.edu/~keisler/foundations.pdf especially p. 34.
In the above I have made a slight abuse of notation by writing $$dy=y(x+dx)-y(x)$$ instead of the more complete and correct $$dy=\text{st}\left( \frac{y(x+dx)-y(x)}{dx} \right) dx$$ Dividing by $dx$, taking the standard part, $\text{st}$, and multiplying by $dx$ is what removes the products of infinitesimals.
I thought that it was not a major abuse of notation since in other contexts $dx\ne 0$ and $dx^2=0$ is a defining property of infinitesimals, but these basic properties of infinitesimals and differentials may not be understood by all readers.