What is the intuition behind the Freudenthal suspension theorem?
Maybe this differential topologic way of thinking the Freudenthal suspension is much more intuitive. By Pontrjagin's contruction you can identify $\pi_{n+k}(S^n)$ with equivalence classes of framed submanifolds $(N,\nu)$ of $S^{n+k}$. The image of this class under the Freudenthal suspension homomorphism is just the framed submanifold $(N,\tilde{\nu})$, where we identify $S^{n+k}$ with the equator of $S^{n+k+1}$ and the frame $\tilde{\nu}$ is obtained by $\nu$ just "adding" to $\nu$ the canonical normal frame of $S^{n+k}$ inside $S^{n+k+1}$. The fact that the map is an isomorphism for $n>k+1$ can now be achieved by general position arguments.
There are two proofs I particularly like:
A Morse-theoretic proof, probably due to Bott, can be found in Milnors book. Idea: consider the space of all paths on $S^n$ from the north to the south pole, which is homotopy equivalent to $\Omega s^n$. There is the energy function on this space. One does Morse theory: critical points correspond to geodesics (they are not non-degenerate, but Bott proved that a good deal of Morse theory works nevertheless). The set of absolute minima is the space of minimal geodesics connecting the poles. This is homeomorphic to $S^{n-1}$. All other critical points have index at least (rouhgly) $2n$. Therefore, the inclusion of the set of absolute minima into the whole space is $2n$-connected.
A spectral sequence proof (see Kirk, Davis, Lectures on Algebraic topology). Consider the homology Leray-Serre spectral sequence of the path-loop fibration $\Omega S^n \to PS^n \to S^n$. The total space $P S^n$ is contractible. Look at the shape of the spectral sequence; you'll see that for $k \leq 2n$ (again, only a rough estimate), all differentials out of the $E_{k,0}$-slot are zero, except of the last one, which gives an isomorphism $H_k (S^n)=E_{k,0}^{2} \to E_{0,k-1}^{2} = H_{k-1} (\Omega S^n)$. It is credible (but nontrivial to prove, this uses the transgression theorem) that this isomorphism is the same as the composition $H_k (S^n) \cong H_{k-1} (S^{n-1}) \to H_{k-1} (\Omega S^n)$ of the suspension isomorphism and the natural map $S^{n-1} \to \Omega S^n$. Thus the natural map is a homology isomorphism in a range of degrees, and by the Hurewicz theorem, this holds for homotopy groups as well.
The Freudenthal theorem is really a special case of the phenomenon called "homotopy excision" aka the Blakers-Massey triad theorem. The idea is that one has an inclusion $$ (C_-X,X) \to (\Sigma X,C_+X) $$ given by gluing in $C_+X$, where $C_\pm X$ are copies of the cone on $X$. The Blakers-Massey theorem tells us that this map is $(2r+1)$-connected, where $r$ is the connectivity of $X$. Furthermore, the inclusion induces the suspension homomorphism on homotopy groups.
There are a variety of proofs of the homotopy excision theorem. In the case at hand, possibly the easiest proof is given in the paper cited by Jeff Strom in answering my question: https://mathoverflow.net/questions/54169.
By the way, in the dual case one has the map $\Sigma \Omega X \to X$. It is somewhat easier to check that the latter is $(2r+1)$-connected. Its homotopy fiber can be identified with $\Sigma (\Omega X) \wedge (\Omega X)$ which is $(2r)$-connected. So the map $\Sigma \Omega X \to X$ is $(2r+1)$-connected.