Let $v:SO(3)\to \mathbb{R}^3$ and $e_1\in\mathbb{R}^3$ and consider $$d\big(R\mapsto [v(R)\cdot e_1]e_1\big) (R)(\omega)$$
I would like to understand if this differential, evaluated at point $R\in SO(3)$ in the direction $\omega\in\mathbb{R}^3$, is equal to
$$\big(d(v)(R)(\omega) \cdot e_1\big)e_1$$ and if so, what properties are involved.
My understanding: $\psi:R\mapsto [v(R)\cdot e_1]e_1$ is equal to $\phi_2\circ\phi_1$ with:
$$\phi_1(R) = v(R),\quad \phi_2(v)= (v\cdot e_1)e_1$$
so
\begin{align}d\psi(R)(\omega) &= d(\phi_2\circ\phi_1)(R)(\omega) = d\phi_2(\phi_1(R)) d\phi_1 (R)(\omega) \\ &= \big(\alpha\mapsto (\alpha\cdot e_1)e_1\big) dv(R)(\omega) \\ &=(dv(R)(\omega)\cdot e_1)e_1 \end{align}
Is that correct? Is this how you would have tackled it?
Let $f\colon SO(3)\to\mathbb{R}^3$ be defined as $f(R)=\langle v(R), e_1\rangle e_1$. Notice that $f=L\circ v$, where $L(u)=\langle u, e_1\rangle e_1$ is linear (and thus satisfies $dL(x)=L$ for $x\in\mathbb{R}^3$).
We have then \begin{equation*} \begin{split} df(R) & = d(L\circ v)(R) \\ & =dL(v(R))\circ dv(R) \\ & = L\circ dv(R) \end{split} \end{equation*} so that $df(R)(\omega)=\langle dv(R)(\omega),e_1\rangle e_1$ for every $\omega\in T_R SO(3)$.
Remark: For each $R\in SO(3)$, the differential $df(R)$ is a linear map from $T_R SO(3)$ to $T_{f(R)}\mathbb{R}^3\cong\mathbb{R}^3$, so that you cannot calculate it in a $\omega\in\mathbb{R}^3$ as you wrote, you must take $\omega\in T_{R} SO(3)$ instead.