It seems to be geometrically clear what the "pinching map" $\mu : S^n \to S^n \vee S^n$ looks like, but we have to be very precise.
Let us assume that the basepoint $*$ of $S^n$ lies in the equatorial sphere $S^{n-1}_0 = S^{n-1} \times \{0\} \subset S^n$. We may simply take $* = (1,0,\dots,0)$. Then the quotient $S^n /S^{n-1}_0 $ is certainly homeomorphic to the wedge of two copies of $S^n$, but this does not specify a concrete $\mu$.
Let us first construct a homeomorphism $D^n/S^{n-1} \to S^n$. The construction is technically non-trivial; it is a slight modification of that presented in my answer to Two topology questions regarding quotient $D^n/S^{n-1}$ and homotopy $S^{n-1} \to S^{n} - \{ a,b\}$. We write $S^n = \{ (r,x) \in \mathbb{R} \times \mathbb{R}^n \mid r^2 + \lVert x \rVert^2 = 1 \}$, where $\lVert - \rVert$ denotes the Euclidean norm.
Define
$$p : D^n \to S^n, p(x) =
\begin{cases}
\left(2\lVert x \rVert -1, \dfrac{\sqrt{1 - (2\lVert x \rVert -1)^2}}{\lVert x \rVert}x \right) & x \ne 0 \\
(-1,0) & x = 0
\end{cases}
$$
This is well-defined because $(2\lVert x \rVert -1) \in [-1,1]$ and
$$ (2\lVert x \rVert-1)^2 + \left\lVert \dfrac{\sqrt{1 - (2\lVert x \rVert -1)^2}}{\lVert x \rVert}x \right\rVert^2= (2\lVert x \rVert -1)^2 + 1 - (2\lVert x \rVert -1)^2 = 1 .$$
$p$ is continuous because for $x \to 0$ we get $\left\lVert \dfrac{\sqrt{1 - (2\lVert x \rVert -1)^2}}{\lVert x \rVert}x \right\rVert = \sqrt{1 - (2\lVert x \rVert -1)^2} \to 0$ and $2\lVert x \rVert -1 \to -1$.
For $x \in S^{n-1}$ we have $p(x) = (1,0) = *$.
Next define
$$j : S^n \setminus \{*\} \to D^n \setminus S^{n-1}, j(r,y) =
\begin{cases}
\dfrac{1 + r}{2\sqrt{1- r^2}}y & y \ne (-1,0) \\
0 & y = (-1,0)
\end{cases}
$$
This is well-defined because $-1 < r < 1$ for $(r,y) \in S^n \setminus \{*, (-1,0)\}$ and
$$\left\lVert \dfrac{1 + r}{2\sqrt{1- r^2}}y \right\rVert = \dfrac{1 + r}{2\sqrt{1- r^2}}\left\lVert y \right\rVert = \dfrac{1 + r}{2} < 1 .$$
It is easily verified that $p(j(r,y)) = (r,y)$ for all $(r,y)$ and $j(p(x)) = x$ for all $x \in D^n \setminus S^{n-1}$. Hence $p$ maps $D^n \setminus S^{n-1}$ bijectively onto $S^n \setminus \{*\}$.
$D^n , S^n$ are compact Hausdorff, hence $p$ is a closed map and thus a quotient map. By the above considerations there exists a unique function $h : D^n/S^{n-1} \to S^n$ such that $h \circ \pi = p$, where $\pi : D^n \to D^n/S^{n-1}$ is the standard quotient map. By construction it is a bijection. By the universal property of quotient maps $p$ and $\pi$ both $h$ and $h^{-1}$ are continuous, i.e. $h$ is a homeomorphism.
Geometrically $h$ can be visualized as follows: Each sphere $\Sigma^{n-1}_r = \{ x \in D^n \mid \lVert x \rVert = r \} \subset D^n$ with $0 < r < 1$ is mapped to the sphere $S^{n-1}_r = S^n \cap (\{2r-1\} \times \mathbb R^n) \subset S^n$, $S^{n-1}$ goes to $*$ and $0$ to $(-1,0)$. That is, $D^n$ is slipped like a rubber glove over $S^n$ beginning from the left (first coordinate $-1$) to the right (first coordinate $1$).
The equatorial ball $D^{n-1}_0 = D^{n-1} \times \{0\} \subset D^n$ is mapped by $h$ onto the equatorial sphere $S^{n-1}_0$. The other level balls $D^{n-1}_t = D^n \cap (\mathbb R^{n-1} \times \{r\})$ are mapped to "inclined subspheres" of $S^n$ which all meet in $*$.
Let us next observe that in definition 1 we can replace $I$ by $J = [-1,1]$ and obviously obtain an isomorphic group, i.e. we may take
$$\pi_n(X,x_0) = [(J^n,\partial J^n),(X,x_0)]$$
with addition defined by $ [f] +[g)] = [f+g]$, where
$$f+g: J^n \to X, (f+g)(x_1,\ldots,x_n) = \begin{cases}f(2x_1+1,x_2,\ldots,x_n) & x_1 \le 0 \\ g(2x_1-1,x_2,\ldots,x_n) & x_1 \ge 0 \end{cases}$$
There is an obvious homeomorphism $\phi : J^n \to D^n$; see my answer to $(D^n\times I,D^n \times 0)$ and $(D^n \times I, D^n \times 0 \cup \partial D^n \times I)$ are homeomorphic (note that $J^n$ and $D^n$ are the closed unit balls with respect to the maximum-norm $\lVert - \rVert_\infty$ and the Euclidean norm). It maps $J^{n-1}_0 = J^{n-1}\times \{0\}$ to $D^{n-1}_0$.
Consider the quotient map $q = p \circ \phi : J^n \to S^n$. Split $J^n$ into the two cuboids $J^n_\pm = \{(x_1,\ldots, x_n) \mid \pm x_1 \ge 0\}$. Then $q$ induces quotient maps $q_\pm : J^n_\pm \to S^n, q_\pm(x_1,\ldots, x_n) = q(2x_1 \mp 1, x_2,\ldots, x_n)$. Since they agree on $J^{n-1}_0$, they can be pasted to a map $Q : J^n \to S^n \vee S^n$. By construction $Q(\partial J^n) = *$, thus $Q$ induces $\mu : S^n \to S^n \vee S^n$. This is the desired explicit pinching map. By construction $\mu(S^{n-1}_0) = \{*\}$.
It is now a routine exercise to verify that your bijection $\mathcal J$ is a group homomorphism.
First of all, the abstract nonsense: The group structure on homotopy groups comes from a cogroup structure on the spheres $S^n,\,n\ge1$ (in the category of pointed topological spaces), also called a co-H space. For a (pointed homotopy class of a) pointed map $f\colon S^k\rightarrow S^n$ to induce a group homomorphism after applying $[-,X]_{\ast}$ for any space $X$ is, by the Yoneda lemma, equivalent to asking that the map $f$ is a cogroup homomorphism, also called a co-H map.
Now, for any pointed space $X$, the pointed suspension $\Sigma X$ has a natural co-H space structure. For $X=S^n$, this is precisely the usual co-H space structure on $\Sigma X=S^{n+1}$ (or perhaps the opposite one, depending on chosen orientation). You are correct to observe that, in general, if $f\colon X\rightarrow Y$ is a pointed map, the suspended map $\Sigma f\colon\Sigma X\rightarrow\Sigma Y$ is a co-H map with respect to these structures.
In particular, the calculation of $\pi_n(S^n)=\mathbb{Z}$ for $n\ge1$ and the notion of degree imply that any pointed map $S^n\rightarrow S^n$ for $n\ge2$ is (up to pointed homotopy) the suspension of the pointed map $S^{n-1}\rightarrow S^{n-1}$ of the same degree, hence a co-H map. For pointed maps $S^1\rightarrow S^1$, the only evident pointed suspensions are the constant map and the identity. The other maps $S^1\rightarrow S^1$ are in fact not co-H maps, for the degree $n$ map $f_n\colon S^1\rightarrow S^1$ induces the $n$-th power map $[f_n]^{\ast}\colon\pi_1(X)\rightarrow\pi_1(X),\,[\gamma]\mapsto[\gamma]^n$. This is not generally a group homomorphism, as one can check with the universal example $X=S^1\vee S^1$.
The general case of pointed maps $S^n\rightarrow S^m$ is hard. If $n<m$, any such map is nullhomotopic and trivially a co-H map. If $n=m$, this has been taken care of, so assume $n>m$. The Freudenthal suspension theorem implies that the suspension map $\Sigma\colon\pi_{i-1}(S^{n-1})\rightarrow\pi_i(S^n)$ is surjective for $i\le 2n-2$, so any map $S^i\rightarrow S^n$ with $i\le 2n-2$ is a (pointed) suspension and hence a co-H map.
It is not too hard to prove that the (pointed) mapping cone of a co-H map between co-H spaces is a co-H space. This allows us to construct non-co-H maps immediately outisde the Freudenthal suspension range. For $n$ even, the Hopf invariant is a non-trivial homomorphism $H\colon\pi_{2n-1}(S^n)\rightarrow\mathbb{Z}$ and the adjunction space obtained by attaching a cell along a map with non-trivial Hopf invariant has non-trivial cup products in positive degrees (by definition), yet that cannot happen for a co-H space (exercise). Thus, such maps are not co-H maps, e.g. the Whitehead product of $\mathrm{id}_{S^n}$ with itself.
There are, in fact, also co-H maps between spheres that are not suspensions, e.g. a map $S^{2p}\rightarrow S^3$ representing an element of order $p$ in $\pi_{2p}(S^3)$, where $p$ is an odd prime. This, among other things, is due to Bernstein and Hilton in "Category and Generalized Hopf Invariants". You can also read Martin Arkowitz' article "Co-H-Spaces" published in the "Handbook of Algebraic Topology" for a neat overview. In fact, the co-H maps between spheres are completely characterized by Bernstein and Hilton in terms of these "generalized Hopf invariants", although they are quite involved.
Best Answer
You probably know how the inverse looks for $n=1$: the inverse of a loop $l(t)$ is the same loop but traversed in the opposite direction: $l^{-1}(t) = l(1-t)$. This way, the loop $l l^{-1}$ is homotopic to a constant loop: one can go from $l(0)$ to $l(1)$ and back and this will be $l l^{-1}$, or one can go from $l(0)$ to $l(\tau)$ only and then go back, for some $\tau \in [0,1].$ For $\tau = 1$ gives $l l^{-1}$ and for $\tau = 0$ gives the the loop constantly equal to $l(0)$.
Now, the pinch map interpretation is the same as the composition of loops, except you add some extra dimensions, in which you don't really do anything. Composing two loops is making a circle into the figure "8" and then mapping each of the little circles by each of the loop maps. In higher dimensions, you just perform this operation of every circular section of the sphere. This should give you an idea of how the inverse is supposed to look.