In my script we took the following identity:
$$\mathcal{R}_{\vec e_y}(-d\alpha')\mathcal{R}_{\vec e_x}(d\alpha)\mathcal{R}_{\vec e_y}(d\alpha')\mathcal{R}_{\vec e_x}(-d\alpha)= \mathcal{R}_{\vec e_z}(d\alpha d\alpha')$$
I'm trying to understand the proof, which goes as this:
We apply these 4 infinitesimal rotations on a vector $\vec{OM}$ to find the resulting vector $\vec {OM}'$. Now for a single rotation the following is valid:
$$\mathcal{R}_{\vec u}(d\alpha)\vec{OM}= \vec{OM}+d\alpha \vec u\times \vec{OM}$$.
It says: It can be immediately seen that if $d\alpha=0$ the left side is identity. The vector $\vec{OM}'-\vec{OM}$ must therefore be proportional to $d\alpha$.For analogous reasons, it must also be proportional to $d\alpha'$.Consequently the difference $\vec{OM}'-\vec{OM}$ is proportional to $d\alpha d\alpha'$.
So, I applied the 2nd formula to each of the 4 rotations, and in the end if you equalize either one from $d\alpha$ or $d\alpha'$ with zero, in both cases you get $\vec{OM}=\vec{OM}$, which is to be expected. So, I don't understand where do we see this proportionality ?
You can surely not get anywhere when you set $d\alpha=d\alpha'=0$ because this trivially leads to $\vec{OM}'=\vec{OM}$.
To get you over the step you have trouble with:
You have to set $d\alpha$ and $d\alpha'$ to zero separately.
When $d\alpha=0$ then the left hand side of the first equation collapses to the identity matrix: $$ \mathcal{R}_{\vec e_y}(-d\alpha')\mathcal{R}_{\vec e_y}(d\alpha')=I\,. $$ Likewise, when $d\alpha'=0\,:$ $$ \mathcal{R}_{\vec e_x}(d\alpha)\mathcal{R}_{\vec e_x}(-d\alpha)= I\,. $$ This means that this left hand side aplied to a vector $\vec{OM}$ $$\tag{1} \mathcal{R}_{\vec e_y}(-d\alpha')\mathcal{R}_{\vec e_x}(d\alpha)\mathcal{R}_{\vec e_y}(d\alpha')\mathcal{R}_{\vec e_x}(-d\alpha)\vec{OM} $$ cannot be of the form $$ \vec{OM}+\vec{V}\,d\alpha+\vec{W}\,d\alpha\,. $$ with nonzero vectors $\vec{V},\vec{W}$. Instead, (1) must be of the form $$ \vec{OM}+\vec{V}\,d\alpha\,d\alpha\,. $$