The following is an excerpt from J. Lee:s "Introduction to Topological Manifolds":
It is not obvious to me that $[f] = [f_1] \ast \ldots \ast [f_n]$ (see the last paragraph). Does one need to construct an explicit path-homotopy to see this? I feel like it is "intuitively obvious", but since J. Lee seem to think this does not warrant any more detail, I suppose it must follow from some theorem covered earlier. I'd be happy for clarification.
First note that
Let $f_k = f \mid_{[(k-1)/n,k/n]}$ reparametrized on the unit interval
is a bit sloppy because it uses the equal sign for two maps defined on different intervals. It would be better to say something like
Let $f_k$ be the path obtained from $f \mid_{[(k-1)/n,k/n]}$ by compressing the unit interval linearly to $[(k-1)/n,k/n]$
The precise meaning is this:
Let $\phi_k : I \to [\frac{k-1}{n},\frac k n], \phi_k(t) = \frac{k-1}{n} + \frac{t}{n} = \frac{k-1 +t}{n}$. Then $f_k = f \mid_{[\frac{k-1}{n},\frac k n]} \circ \phi_k$.
In my answer to Proving Munkres Theorem 51.3 I showed that $[f_1] \ast \ldots \ast [f_n]$ can be represented as follows:
Take the equidistant partition $\mathbf a = (a_0,a_1, \ldots,a_n)$ of $I$ with $a_k = \frac k n$ for $k = 0,\ldots, n$. Then $[f_1] \ast \ldots \ast [f_n]$ is represented by the path $f_\mathbf a$ given by $$f_\mathbf a(t) = f_k\left(\frac{t-a_{k-1}}{a_k -a_{k-1}}\right) = f_k(nt- (k-1)) \text{ for } t \in [a_{k-1},a_k] = [\frac{k-1}{n},\frac k n] .$$ But $f_k(nt -(k-1)) = f(\phi_k(nt- (k-1))) = f(t)$. This shows that $f = f_\mathbf a$, i.e. $$[f] = [f_1] \ast \ldots \ast [f_n] .$$
Here is a partial answer, where I in the end fail to make it overlap globally, but it seems to me that there should be a way to define it consistently globally.
Also see the link provided by Paul Frost in the comments, which seems interesting.
We let $f:I \to M$ be a loop based at $p$ with image in a manifold $M$, and make the restriction $f|_{\left[\frac{(k-1)}{n},\frac{k}{n}\right]}$. Then I believe $f_k = f\left(\frac{(k-1)}{n}(1-t)+\frac{tk}{n}\right)$ is a reparametrization of $f|_{\left[\frac{(k-1)}{n},\frac{k}{n}\right]}$; to see this, note that $\gamma(t) = \frac{(k-1)}{n}(1-t)+\frac{tk}{n}$ has derivative $\frac{1}{n}$ and $I$ is convex, so $f_k:I \to M$ will take all values at $f$ inside $\left[\frac{(k-1)}{n},\frac{k}{n}\right]$ (we are using that $\gamma$ has domain $I$ which is a closed interval, so by the intermediate value theorem, takes on every value between $\left[\frac{(k-1)}{n},\frac{k}{n}\right]$ ).
We can now define a path-homotopy $f_k \sim f$ as follows: \begin{align*} H_k(s,t) &= f\left(st+\left(\left(\frac{k-1}{n}\right)(1-s)+\frac{sk}{n}\right)(1-t)\right),\qquad(1) \end{align*} for $k = 1,\ldots,n$. One checks that this is indeed a path-homotopy, i.e. that \begin{align*} H(s,0) &= f_k(s)\\ H(s,1) &= f(s)\\ H(0,t) &= p\\ H(1,t) &= p. \end{align*}
This seem to work (please correct me otherwise).
We let $f:I \to M$ be a loop based at $p$ with image in a manifold $M$, and make the restriction $f|_{\left[\frac{(k-1)}{n},\frac{k}{n}\right]}$. Then I believe $f_k = f\left(\frac{(k-1)}{n}(1-t)+\frac{tk}{n}\right)$ is a reparametrization of $f|_{\left[\frac{(k-1)}{n},\frac{k}{n}\right]}$; to see this, note that $\gamma(t) = \frac{(k-1)}{n}(1-t)+\frac{tk}{n}$ has derivative $\frac{1}{n}$ and $I$ is convex, so $f_k:I \to M$ will take all values at $f$ inside $\left[\frac{(k-1)}{n},\frac{k}{n}\right]$ (we are using that $\gamma$ has domain $I$ which is a closed interval, so by the intermediate value theorem, takes on every value between $\left[\frac{(k-1)}{n},\frac{k}{n}\right]$ ).
We then define $H(s,t)$ as follows: \begin{align*} (2) \qquad H(s,0) &= \begin{cases} H_1(ns,0), 0 \leqslant s \leqslant \frac{1}{n}\\ H_2(ns-1,0), \frac{1}{n} \leqslant s \leqslant \frac{2}{n}\\ \vdots \\ H_n(ns-(n-1),0), \frac{n-1}{n} \leqslant s \leqslant \frac{n}{n} = 1 \end{cases} \end{align*}
We then see that $H(s,0) = f_1 \ast \ldots \ast f_n$.
Furthermore, we define \begin{align} \begin{aligned} (3) \qquad H(s,1) &= \begin{cases} H_1(s,1), 0 \leqslant s \leqslant \frac{1}{n}\\ H_2(s,1), \frac{1}{n} \leqslant s \leqslant \frac{2}{n}\\ \vdots \\ H_n(s,1), \frac{n-1}{n} \leqslant s \leqslant \frac{n}{n} = 1 \end{cases} \end{aligned} \end{align}
One then finds that $H(s,1) = f(s)$. It remains to see how to define $H(s,t)$ for $0 < t < 1$. $(2)$, $(3)$ gives us a hint.
From $(2)$, $(3)$ one would perhaps naively define it globally as \begin{align*} H(s,t) &= \begin{cases} H_1(st+(1-t)(ns),t), 0 \leqslant s \leqslant \frac{1}{n}, 0 \leqslant t \leqslant 1\\ H_2(st+(1-t)(ns-1),t), \frac{1}{n} \leqslant s \leqslant \frac{2}{n},0 \leqslant t \leqslant 1\\ \vdots \\ H_n(st+(1-t)(ns-(n-1)),t), \frac{n-1}{n} \leqslant s \leqslant \frac{n}{n} = 1, 0 \leqslant t \leqslant 1 \end{cases} \end{align*}
but (!) this does not work, because $H(s,t)$ as defined does not seamlessly overlap on $s = \frac{k}{n}$ for $H_{k},H_{k+1}$.