= Solution
Write $T=\varepsilon t$ and expand
$$
\frac1R=\frac1{R_c+\varepsilon^2}
=-f'(0)-\varepsilon^2f'(0)^2+O(\varepsilon^4).
$$
At order $\varepsilon^2$, the assumptions $f''(0)=0$ and the orthogonality normalization leave a homogeneous equation for $u_2$ with no forcing, hence $\boxed{u_2=0}$.
At order $\varepsilon^3$, project the equation onto the null mode $\sin z$. Since
$$
\frac{\int_0^\pi\sin^4z\,dz}{\int_0^\pi\sin^2z\,dz}=\frac34,
$$
the <Fredholm solvability condition> obtained directly from the definitions printed in the question is
$$
\boxed{
\frac{d^2A}{dT^2}
=f'(0)^2A-\frac18f'''(0)A^3}.
$$
The paper asks for $A/f'(0)^2$ in place of $f'(0)^2A$, but that coefficient is incompatible with $R_c=-1/f'(0)$ and $\varepsilon^2=R-R_c$: differentiating $1/R$ at $R_c$ gives $-1/R_c^2=-f'(0)^2$. Thus the displayed target appears to contain a reciprocal typo. It would agree with the expansion only under a correspondingly rescaled definition of the small parameter.
The imposed orthogonality of every $u_j$, $j\geq2$, removes the freedom to transfer a multiple of $\sin z$ between $A$ and the higher-order terms.
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