= Solution
The <curvature form of a connection> is the endomorphism-valued <differential two-form> defined by
$$
R^\nabla(X,Y)s=\nabla_X\nabla_Ys-\nabla_Y\nabla_Xs-\nabla_{[X,Y]}s.
$$
It is alternating in $X,Y$. To see its <tensoriality>, replacing $X$ by $fX$ produces an extra $-Y(f)\nabla_Xs$ from the second term and an extra $+Y(f)\nabla_Xs$ from $[fX,Y]=f[X,Y]-Y(f)X$, so these cancel. Alternation gives linearity over <smooth functions> in $Y$ too. Replacing $s$ by $fs$ produces the additional coefficient $(X(Yf)-Y(Xf)-[X,Y]f)s=0$; all terms involving one derivative of $s$ also cancel. Thus $R^\nabla$ belongs to $\Omega^2(M;\operatorname{End}E)$.
In the coefficient-column convention of part (b), expanding the definition gives
$$
R^\nabla(X,Y)(eu)=e\Bigl(
X(\Omega(Y))-Y(\Omega(X))-\Omega([X,Y])
+\Omega(X)\Omega(Y)-\Omega(Y)\Omega(X)\Bigr)u.
$$
The first three terms are $d\Omega(X,Y)$, and the last two are $(\Omega\wedge\Omega)(X,Y)$. This proves the <Cartan curvature matrix equation> $F=d\Omega+\Omega\wedge\Omega$, with matrix multiplication combined with the <wedge product of differential forms>.
The <pullback connection> has matrix $\Omega'=\phi^*\Omega$. Since the <pullback of a differential form> commutes with the <exterior derivative> and the <wedge product of differential forms>, its <curvature form of a connection> has matrix
$$
F'=d(\phi^*\Omega)+(\phi^*\Omega)\wedge(\phi^*\Omega)
=\phi^*(d\Omega+\Omega\wedge\Omega)=\phi^*F.
$$
Changes of <frame of a vector bundle> conjugate both sides by the same pulled-back transition matrix, so this identity is intrinsic. In fibre notation the answer is
$$
\boxed{R^{\nabla'}_p(u,v)=R^\nabla_{\phi(p)}(d\phi_pu,d\phi_pv),
\qquad u,v\in T_pM'.}
$$
Both sides act on the same fibre $E_{\phi(p)}$. No claim that $d\phi$ preserves <Lie brackets of vector fields> for an arbitrary map is needed.
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