The example I have in mind of a torsion element in fundamental group is what happens on the projective plane: there is a cell $D^2$, but its boundary doubly covers a copy of $S^1$. The latter has then order two in the homotopy group. Let me call this a "wrapped 2-disk" sitting inside the projective plane.
In general, a torsion element $\gamma \in \pi_k(S^n) $ can be thought as a commutative diagrams
$$ \begin{array}{ccc} \partial D^{k+1} & \rightarrow & D^{k+1} \\ \downarrow & & \downarrow \\ S^k & \rightarrow & S^n \end{array} $$
For the left vertical arrow having finite non zero index, and this is a wrapped $(k+1) $ disk. It is known that even for low indices torsion element exists; do you know an explicit representative? I suspect it could be symmetric enough.
Any geometric idea on how this "finite wrapping" between different dimensions should arise is welcomed.
With the classical description of homotopy groups of spheres, the best answer is probably "It appears because there is no reason for it not to." If I recall correctly, every finite, simply-connected, and noncontractible CW complex has torsion in its homotopy groups - this is a result of Serre.
If you allow different descriptions of homotopy groups of spheres, then there can be more interesting explanations. In the comments Кряжев Арсений, alludes to framed manifolds. It is a result of Thom, I believe, that says the stable homotopy groups of spheres are isomorphic to the cobordism ring of framed manifolds. Here stable refers to a specific subcollection of the homotopy groups of spheres (or alternatively the colimit of the homotopy groups under the suspension homomorphism). Framed means manifolds equipped with a trivialization of their stable normal bundle, and cobordism means take these decorated manifolds up to the equivalence relation generated by $M \simeq N$ if there is a decorated manifold with boundary $W$ so that $\partial W = M \cup N$.
Then under this isomorphism, the suspension of the Hopf map is represented by the circle and instead of the canonical framing of its normal bundle in $\mathbb{R}^3$, we twist it once as we go along the circle. This can't be extended along a manifold with boundary $S^1$, so it gives a nontrivial element of the stable homotopy groups, but two copies of this bound a cylinder the framing can be extended along, so the element has order 2.
There is a another reasonable description I like of higher homotopy groups in the homotopy groups of spheres, and that is that it is required by the existence of cohomology operations. It's a classical result that the spaces with a single nontrivial homotopy group in degree n represent nth cohomology with coefficients in that homotopy group. By the Yoneda lemma, the cohomology of these spaces represents natural transformations between cohomology groups. There are many calculations of these natural transformations, so called cohomology operations, and they surely exist.
Since the cohomology of $S^n$ is nontrivial only in degree n, the existence of cohomology operations implies that this cannot represent nth singular cohomology, so it must have multiple nontrivial homotopy groups. Additionally, it is not hard to prove that if n is odd, $S^n$ has no higher homotopy groups with nontorsion in them. So we see in this case, that torsion is forced to exist as a result of cohomology operations existing.
Of course there are direct ways to show many nontorsion elements, and essentially all the computational ways will leave you with the feeling "This is just here by chance." And they really are, the homotopy groups of spheres are incredibly chaotic, and they are definitely not understood through the classical definition of maps of spheres that bound a disk. Topologists try to abstract away until we are very far from this original definition, using it only when necessary. If you have a geometric mind you will like Кряжев Арсений's description, and if you have a more algebraic mind you will like things like the Adams Spectral sequence. Both explain why torsion arises, and both are very far from the original definition.