A harmonic function cannot have a local minimum unless it is constant. One way to see this is variational: Suppose $u(x_0)=m$ is a strict local minimum and consider $\tilde u = \max(u,m+\epsilon)$ in a neighborhood of $x_0$ (and $\tilde u=u$ elsewhere). The function $\tilde u$ is a local perturbation of $u$ and has strictly smaller Dirichlet energy (since it is constant on part of the domain and agrees with $u$ otherwise). This contradicts the minimizing property of harmonic functions.
Another way to see this without calculation is to recall that the Laplacian is the divergence of the gradient. If $u$ has a strict local minimum at $x_0$, then in some spherical neighborhood of $x_0$ the gradient points away from $x_0$. Thus, its flux across the boundary of this neighborhood is nonzero, contradicting the divergence theorem.
Yet another interpretation: the fundamental solution comes from a point source. The gradient field has no sinks or sources apart from the singularities. A sphere is a compact manifold: nowhere to escape. So there must be a sink somewhere. For the symmetric solution they consider, the sink will be at the antipodal point of the source. E.g., the function is $-\infty$ at one pole, $+\infty$ at the other, and zero along the equator.
And yet another justification for the above is the reflection principle: if the harmonic function is zero on the equator of a sphere, then it is odd in the sense that $u(Rx)=-u(x)$ where $R$ is the reflection about the equator. So the singularity is reflected too.
Things are different on noncompact manifolds such as $\mathbb R^n$, because the gradient flow can try to escape to infinity. It escapes without trouble when $n\ge 3$. But when $n=2$, the exit to infinity is so narrow that a stampede results: the fundamental solution (logarithm) blows up at infinity.
