Let $X$ be an $H$-space, i.e. there is a multiplication $\mu\colon X\times X\to X$ continuous with an identity $e$ such that $(x\mapsto \mu(e,x))\simeq (x\mapsto\mu(x,e))\simeq \text{id}_X$.
I want to show that the addition of two maps $f,g\in\pi_n(X,x_0)$ may be given by $(f+g)(x)=\mu(f(x),g(x))$, but I am unsure of what to do exactly.
This is problem 4.1.3 in Hatcher's Algebraic Topology book.
This depends on a results called the "interchange law" or the "Eckman-Hilton argument".
Let $S $ be a set with two monoid structures $\circ_1, ~ \circ_2 $ each of which is a morphism for the other. Then the two monoid structures coincide and are abelian.
It is crucial for this that a morphism of monoids preserves the identities.
Here is a sketch of the argument.
The compatibility of the two structures is expressed by saying that the matrix composition $$\begin{bmatrix}x&y \\ z&w \end{bmatrix}$$ where vertically gives $\circ_1$ and horizontally gives $\circ_2$, has only one result, which linear notation says $$(x \circ_1 z)\circ_2(y \circ_1 w)= (x \circ_2y)\circ_1 (z \circ_2 w).$$ Now you make substitutions to obtain $e_1=e_2$, i.e. equality of the identities, which you now write as $e$, and further substitutions to obtain $x \circ_1 y= x\circ_2y$, which we now write as $x \circ y$, and finally, with another sustitution, obtain $x\circ y= y \circ x$.
Note that this argument does not work if one composition is a a monoid and the other is a semigroup, or if one is a category, so with many identities. So "higher dimensional groups" are abelian groups, but higher dimensional groupoids can be much more complicated than groupoids, and so model more geometry.