There is clearly a typo in this question, since the $J_{0} \subseteq J$ part is never used. Moreover, the following hypothesis does not seem at all relevant in proving the result:
Since you can simply choose $J=I$ and $\theta_i=\neg\phi_i$.
What would be the intended meaning of this question?
There was probably a similar statement about the $\theta_j$ and $J$ that was forgotten.
In fact you can see that the hypothesis is completely symmetric in $(\phi_i: i\in I)$ vs $(\theta_j: j\in J)$. So the intended statement is probably that the same holds for $(\theta_j: j\in J)$, i.e. there exists a finite $J_0\subset J$ such that $\mathcal{M}\models \bigvee_{j\in J}\theta_j \leftrightarrow \bigvee_{j\in J_0}\theta_j$.
However, your later remark is wrong : $\bigvee_{i\in I}\neg \phi_i$ is not, in general, equivalent to $\neg\bigvee_{i\in I}\phi_i$ : the hypothesis does play a role.
The inteded meaning is some sort of compactness result. The idea is that $|I|,|J|<\kappa$ so there can be at most $\kappa$-many symbols from $\mathcal{M}$, and the rest will come from $\mathcal{L}$ itself, so you can use $\kappa$-saturation of $\mathcal{M}$, and find appropriate types that must be realized in $\mathcal{M}$ if the conclusion fails : then you show that they can't be realized, based on the hypothesis, and you're done.