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I think you just need to pick a "product" coordinate neighborhood $U \cong U_1 \times U_2$ inside $\Phi^{-1}(W)$. Knowing that $\Phi^* f = 0$ in that neighborhood allows you to conclude that $f$ is zero on $\Phi(U) \cong U_2$.
In any case, can you also get the bound on $\left( \sum_k 2^{2k} |P_kf(x)|^2 \right)^{1/2}$ directly from $\sum_{k ≤ k_0} 2^{2k} |P_k f(x)|^2 ≲ 2^{2k_0} B(x)^2$ and $\sum_{k > k_0} 2^{2k} |P_k f(x)|^2 ≤ 2^{-2k_0}\sum_{k > k_0} 2^{4k} |P_k f(x)|^2 = 2^{-2k_0} A(x)^2$?
