Let $M = \mathbb S^2 \times \mathbb R$ with the Riemannian product metric $g = g_0 \oplus dr^2$, where $g_0$ is the round metric on $\mathbb S^2$ and $r$ is the unit coordinate on $\mathbb R$. Let $X(\phi,\theta) = (\cos\theta\sin\phi, \sin\theta\sin\phi, \cos\phi)$ be spherical coordinates on $\mathbb S^2$. Suppose $\Pi$ is a plane in $T_pM$ for some $p = (\phi,\theta,r) \in M$ that's tangent to $\mathbb S^2 \times \{r\}$. Intuitively, one would think the sectional curvature of $\Pi$ is $1$ since the geodesics with initial velocities in $\Pi$ are great circles of $\mathbb S^2$.
However, I calculated that $M$ is flat. I found that $g = \sin^2\phi\,d\theta^2 + d\phi^2 + dr^2$, whence one can show that the only nonzero Christoffel symbol of the Levi-Civita connection on $M$ is $\Gamma_{\phi\phi}^\phi = \cot\phi$, whence one can show that the curvature tensor vanishes identically, so $M$ is flat and $\sec(\Pi) = 0$. Is there an intuitive reason behind this? Am I wrong in assuming $\mathbb S^2 \times\{r\} \subset M$ is totally geodesic for each $r$? Or did I make an error in my calculation and $(M,g)$ is not flat after all?