= Solution
For canonical $K$, extract the $\theta^2\bar\theta^2$ component of $\Phi_i^\dagger\Phi_i$. The part depending on the <auxiliary fields> is $F_i^*F_i$. To extract the <F-term> of the <superpotential>, expand around the <complex scalar fields>:
$$
W(\Phi)=W(\phi)+W_i(\sqrt2\theta\psi_i+\theta^2F_i)+\frac12W_{ij}(\sqrt2\theta\psi_i)(\sqrt2\theta\psi_j).
$$
All higher terms contain at least three identical-type <Grassmann variables> and vanish. In the convention $(\theta\psi_i)(\theta\psi_j)=-\tfrac12\theta^2\psi_i\psi_j$, the highest component is $W_iF_i-\tfrac12W_{ij}\psi_i\psi_j$. Therefore
$$
\boxed{\mathcal L_F=\sum_iF_i^*F_i+\sum_i(W_iF_i+\overline{W_i}F_i^*),\qquad\mathcal L_{\rm Yukawa}=-\frac12W_{ij}\psi_i\psi_j+\mathrm{h.c.}}
$$
Here $W_i=\partial W/\partial\phi_i$ and $W_{ij}=\partial_i\partial_jW$. The <Yukawa interaction> is displayed to distinguish it from the terms actually containing $F$. With signature $(+---)$, the other canonical terms are $\partial_\mu\phi_i^*\partial^\mu\phi_i+i\bar\psi_i\bar\sigma^\mu\partial_\mu\psi_i$, up to a <total derivative>.
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