The standard (and probably the simplest) example of a manifold with two non-parabolic ends is obtained by taking two disjoint copies of $\mathbb R^3$ and joining them with a "bridge" (or a "tube"). Then the space of bounded harmonic functions on the resulting manifold $M$ is two-dimensional, which means that the gradient vector fields corresponding to bounded harmonic functions form a one-dimensional space, i.e., all of them are proportional to a certain non-zero field $v$.

Now, instead of just one tube one can take two (or more) tubes, and make sure that the whole construction is symmetric. The field $v$ above must also be symmetric, so that it has to vanish at any fixed point of such a symmetry.

More rigorously, let $g$ be an isometry of $M$ which fixes its ends. Then $g$ also fixes the field $v$. It leads to a contradiction if $g$ also fixes a point $p\in M$ without fixing any non-zero vector in $T_p M$.

For a concrete example take in $\mathbb R^4$ two parallel copies $A$ and $A'$ of $\mathbb R^3$, which we shall identify by the means of the corresponding orthogonal projection. Let $D_1, D_2\subset A$ be two disjoint disks of the same radius, and let $D'_1,D'_2$ be the corresponding disks in $A'$. Remove all of them, join the corresponding boundaries with cylindrical tubes, and make the surgery smooth in the same rotationally invariant way on all boundaries. Then the central symmetry of $A$ whose center $p$ is halfway between the centers of $D_1$ and $D_2$ obviously extends to an isometry of the resulting manifold $M$ which satisfies the above conditions.