Geometric Intuition Behind Metric Tensors?

Question

Given the metric tensor of a sphere in matrix form, $$g_{ij}=\begin{bmatrix} 1 & 0 & 0\\ 0 & r^2 & 0\\ 0 & 0 & r^2\sin^2\theta \end{bmatrix},\qquad x^i=(r,\theta,\phi)$$ I can't seem to understand why the $\textbf{e}_{\phi}$ basis vector given by $\sqrt{g_{\phi\phi}}=r\sin\theta$ is a function of $\theta$ in addition to $r$. If the manifold is spherically symmetric, why wouldn't $g_{\theta\theta}=g_{\phi\phi}$? Additionally, you could replace $r\sin\theta$ with $y$ in Cartesian coordinates, indicating that the magnitude of $\textbf{e}_{\phi}=0$ when $y=0$. I understand how the metric is derived but would appreciate some geometrical intuition.