I want to obtain the following formula:
For any $(X,A,x_0)$ the formula $a+b-a=(\partial a)b$ holds for all $a,b \in \pi_2(X,A,x_0)$.
Here $\partial : \pi_2(X,A,x_0) \rightarrow \pi_1(A,x_0)$ is the usual boundary map and $(\partial a)b$ denotes the action of $\partial a$ on b.
The ‘+’ and ‘-‘ refer to the group operation in the nonabelian group $\pi_2(X,A,x_0)$.
So I know that the formula is obtained by constructing a homotopy from $a+b-a$ to $(\partial a)b$. But I’m unsure on how to do it formally.
Let $I^{2}$ denote the unit square, $I^{1}$ denote the bottom side of $I^{2},$ and $J^{1} = \overline{\partial I^{2} - I^{1}}$ denote the union of the other three sides of $I^{2}.$ Then, $\pi_{2}(X, A, x_{0})$ consists of homotopy classes of maps $(I^{2}, I^{1}, J^{1}) \to (X, A, x_{0}).$
Here's Hatcher's construction of the homotopy between the two, which you can find in the proof of Lemma 4.39 of his algebraic topology book:
We start off with $a + b -a,$ represented in the leftmost image. In the second image, we've replaced $a$ with the map $a',$ which is defined as follows: $$a'(s_{1}, s_{2}) = \begin{cases}\partial a(s_{1}), \; &s_{2} \in [0, \frac{1}{2}], \\ a(s_{1}, 2s_{2}-1), \; &s_{2} \in [\frac{1}{2}, 1].\end{cases}.$$ Note that this is well-defined because when $s_{2} = \frac{1}{2},$ $(s_{1}, 2s_{2}-1) = (s_{1}, 0) \in I^{1},$ so $a(s_{1}, 0)$ agrees with $\partial a(s_{1})$ (because $\partial a$ is just the restriction of $a$ to $I^{1}$).
This formal description is not so important. What's important is represented by the lines in the picture: on each of these straight lines, $a'$ is $\partial a,$ starting with $\partial a(0) = x_{0}$ on the left side of the big rectangle and ending with $\partial a(1) = x_{0}$ on the left of the square labelled $b$. On the top half of the rectangle $a'$ is $a$ (rescaled to fit).
Similarly, in the second picture, $-a$ is replaced with $-a'$, which is $\partial a$ on each of those straight lines on the right, starting with $\partial a(0) = x_{0}$ on the right side of the big rectangle and ending with $\partial a(1) = x_{0}$ on the right of the square labelled $b$.
Note that $a$ is obtained from $a$ by holding $s_{2}$ constant for half a second and then speeding up the $s_{2}$ coordinate two times, so $a'$ is homotopic to $a$. Similarly, $-a'$ is homotopic to $-a$.
As for the middle of the second image, we replace $b$ with $b',$ which is $b(s_{1}, 2s_{2})$ when $s_{2} \in [0, \frac{1}{2}]$ and $x_{0}$ when $s_{2} \in [\frac{1}{2}, 1]$. Since $b(s_{1}, 1) = x_{0}$ (because $(s_{1}, 1) \in J^{1})$), $b'$ is well-defined, and we can see that $b'$ is homotopic to $b.$ So, the map represented by the second image is homotopic to the first.
In the third image, we can replace $x_{0}$ with $a + (-a)$ (making the top row of the rectangle half $a$ and half $-a$), and then start replacing more of the $a$ and $-a$'s area with $\partial a$ line-segments. This is well defined by the same argument that $a', -a'$ are well-defined above. This is homotopic to the second image by a similar argument that $a'$ is homotopic to $a$ (replacing $a$ with $\partial a$ is like holding $s_{2}$ constant for longer).
Finally, in the fourth image, we've replaced all of the $a$ and $-a$ with radial line segments, and on each of these radial line segments we run $\partial a$ (where $\partial a(0)$ is on the outside boundary and $\partial a(1)$ is on the inside boundary of the smaller square). Finally, in the smaller square, we just have $b$ (rescaled). This is homotopic to the map in the third image by a very similar argument. Furthermore, this map in the fourth image is precisely the map $(\partial a)b,$ so we are done.