Is there a "geometric reason" for why the gradient of $f(x,y)=\frac{x}{\sqrt{x^2+y^2}}$ is tangent to the unit circle?

I think you have basically answered the question yourself (I would argue that the fact that $f$ depends only on $\theta$ in the polar coordinates is the right intuition).

Another way to understand the result is the fact that $f(x,y)$ is homogeneous of degree $0$, that is to say, we have $$f(\lambda x, \lambda y) = f(x,y) \tag{1}$$ for $\lambda>0$. Now taking the derivative of (1) with respect to $\lambda$, we obtain $$ x \frac{\partial f}{\partial x}(\lambda x, \lambda y) +y \frac{\partial f}{\partial y}(\lambda x, \lambda y) =0$$ and setting $\lambda =1$, the result follows.