One way to think about the action of $\pi_1(X)$ on the higher homotopy groups of $X$ is to think of it as being induced by the action of $\pi_1(X)$ on the universal cover $\widetilde{X}$, which has the same higher homotopy groups as $X$. We can try to reverse this argument, and to construct the desired space by first constructing its universal cover with the desired homotopy groups, then constructing the desired action of $\pi_1(X)$ on it, and finally quotienting by this action appropriately.
Constructing $\widetilde{X}$ is the easiest part: we can just take it to be a product $\prod_{n \ge 2} B^n \pi_n(X)$ of Eilenberg-MacLane spaces (where by $B^n A$ I mean $K(A, n)$). If you believe that the construction of Eilenberg-MacLane spaces is functorial then any desired action of $\pi_1(X)$ on each $\pi_n(X)$ induces an action on $B^n \pi_n(X)$ and hence we get the desired action of $\pi_1(X)$ on $\widetilde{X}$.
The tricky part now to make sure that quotienting $\widetilde{X}$ by $\pi_1(X)$ actually gives a covering map, so that the quotient $X$ actually has the correct homotopy groups. This is what the Borel construction is for; it's a distinguished way to modify $\widetilde{X}$ in a way that preserves both its (weak) homotopy type and the action of $\pi_1(X)$ on it, but so that the action of $\pi_1(X)$ is free.
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.
Best Answer
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.