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Here is a first attempt:

Let $P_{i,j,k}=\{1_{i,j,k},\ldots,n_{i,j,k},\ldots,\gamma(i,j,k)_{i,j,k}\}$ (so that for each choice of $i$, $j$, and $k$, we have a disjoint set of size $\gamma(i,j,k)$). Let $Q=\{p_k\mid k\in K\}$ (so just $K$, up to relabeling). Let $R=\{\ast\}$. Let $$X=Q\coprod R\coprod_{\substack{i\in I,j\in J\\\k\in K}}P_{i,j,k}.$$ Define $$\Omega_j=\coprod_{i\in I,k\in K}P_{i,j,k}\subset X,$$ and $f_i:X\rightarrow X$ by $$f_{i}(n_{i_0,j_0,k_0})=\begin{cases}p_{k_0}\text{ if }i=i_0\\\ \ast\text{ otherwise}\end{cases}$$ $$f_i(p_k)=\ast\hskip0.3in f_i(\ast)=\ast$$ Thus $$f_i^{-1}(p_k)=\coprod_{j\in J}P_{i,j,k}\subset X,$$ and thus $f_i^{-1}(p_k)\cap \Omega_j=P_{i,j,k}$, so $|f_i^{-1}(p_k)\cap\Omega_j|=\gamma(i,j,k)$.

Unfortunately this doesn't address your size concerns, i.e. the preimage of any element of $X$ being countable. Perhaps this can be fixed by adding more null elements (i.e. more $\ast$'s). I'll leave this as a community wiki, and if anyone sees a way of fixing it they are welcome to edit this.

Here is a second attempt (see edit history for previous version).

For each $t\in\mathbb{N}$, let $$P_{i,j,k,t}=\{1_{i,j,k,t},\ldots,n_{i,j,k,t},\ldots,\gamma(i,j,k)_{i,j,k,t}\}$$ (so that for each choice of $i\in I$, $j\in J$, $k\in K$, and $t\in\mathbb{N}$, we have a disjoint set of size $\gamma(i,j,k)$).

For each $t\in\mathbb{N}$, let $$Q_t=\{a_{k,t}\mid k\in K\}$$ (so for each $t\in\mathbb{N}$, this is just a copy of $K$, up to relabeling). 

Let $$X=\coprod_{t\in\mathbb{N}}\left(Q_t\coprod_{\substack{i\in I,j\in J\\\k\in K}}P_{i,j,k,t}\right).$$ Define $$\Omega_j=\coprod_{i\in I,k\in K}P_{i,j,k,1}\subset X,$$ and $f_i:X\rightarrow X$ by $$f_{i_0}(n_{i,j,k,t})=\begin{cases}a_{k,1}\text{ if }i=i_0,t=1\\\ n_{i,j,k,t+1}\text{ otherwise}\end{cases}$$ $$f_i(a_{k,t})=a_{k,t+1}$$

Thus $$f_{i}^{-1}(n_{i,j,k,t})=\begin{cases}\emptyset\text{ if }t=1,2\\\ \{n_{i,j,k,t-1}\}\text{ if }t>2\end{cases}$$ $$f_i^{-1}(a_{k,t})=\begin{cases}\coprod_{j\in J}P_{i,j,k,1}\text{ if }t=1\\\ \{a_{k,t-1}\}\text{ if }t>1\end{cases}$$ We choose $p_k=a_{k,1}$.

Thus $f_i^{-1}(p_k)\cap \Omega_j=P_{i,j,k,1}$, so $|f_i^{-1}(p_k)\cap\Omega_j|=\gamma(i,j,k)$.

Unfortunately this still doesn't address your size concerns, i.e. the preimage of any element of $X$ being countable, because if $J$ is uncountable then $f_i^{-1}(a_{k,1})$ is uncountable (I added the whole mess with the $t$'s to make the preimages of all the other elements countable). I'll leave this as a community wiki, and if anyone sees a way of fixing it they are welcome to edit this.

Let $P_{i,j,k}=\{1_{i,j,k},\ldots,n_{i,j,k},\ldots,\gamma(i,j,k)_{i,j,k}\}$ (so that for each choice of $i$, $j$, and $k$, we have a disjoint set of size $\gamma(i,j,k)$). Let $Q=\{p_k\mid k\in K\}$ (so just $K$, up to relabeling). Let $R=\{\ast\}$. Let $$X=Q\coprod R\coprod_{\substack{i\in I,j\in J\\\k\in K}}P_{i,j,k}.$$ Define $$\Omega_j=\coprod_{i\in I,k\in K}P_{i,j,k}\subset X,$$ and $$f_{i}(n_{i_0,j_0,k_0})=\begin{cases}p_{k_0}\text{ if }i=i_0\\\ \ast\text{ otherwise}\end{cases}$$ $$f_i(p_k)=\ast\hskip0.3in f_i(\ast)=\ast$$ Thus $$f_i^{-1}(p_k)=\coprod_{j\in J}P_{i,j,k}\subset X,$$ and thus $f_i^{-1}(p_k)\cap \Omega_j=P_{i,j,k}$, so $|f_i^{-1}(p_k)\cap\Omega_j|=\gamma(i,j,k)$.

Here is a first attempt:

Let $P_{i,j,k}=\{1_{i,j,k},\ldots,n_{i,j,k},\ldots,\gamma(i,j,k)_{i,j,k}\}$ (so that for each choice of $i$, $j$, and $k$, we have a disjoint set of size $\gamma(i,j,k)$). Let $Q=\{p_k\mid k\in K\}$ (so just $K$, up to relabeling). Let $R=\{\ast\}$. Let $$X=Q\coprod R\coprod_{\substack{i\in I,j\in J\\\k\in K}}P_{i,j,k}.$$ Define $$\Omega_j=\coprod_{i\in I,k\in K}P_{i,j,k}\subset X,$$ and $f_i:X\rightarrow X$ by $$f_{i}(n_{i_0,j_0,k_0})=\begin{cases}p_{k_0}\text{ if }i=i_0\\\ \ast\text{ otherwise}\end{cases}$$ $$f_i(p_k)=\ast\hskip0.3in f_i(\ast)=\ast$$ Thus $$f_i^{-1}(p_k)=\coprod_{j\in J}P_{i,j,k}\subset X,$$ and thus $f_i^{-1}(p_k)\cap \Omega_j=P_{i,j,k}$, so $|f_i^{-1}(p_k)\cap\Omega_j|=\gamma(i,j,k)$.

Unfortunately this doesn't address your size concerns, i.e. the preimage of any element of $X$ being countable. Perhaps this can be fixed by adding more null elements (i.e. more $\ast$'s). I'll leave this as a community wiki, and if anyone sees a way of fixing it they are welcome to edit this.

Let $P_{i,j,k}=\{1_{i,j,k},\ldots,n_{i,j,k},\ldots,\gamma(i,j,k)_{i,j,k}\}$ (so that we have a distinct set of size $\gamma(i,j,k)$, one for each choice of $i,j,k$). Let $Q=\{p_k\mid k\in K\}$ (so just $K$, up to relabeling). Let $R=\{\ast\}$. Let $$X=Q\coprod R\coprod_{i\in I,j\in J,k\in K}P_{i,j,k}.$$ Define $$\Omega_j=\coprod_{i\in I,k\in K}P_{i,j,k}\subset X,$$ and $$f_{i}(n_{i_0,j_0,k_0})=\begin{cases}p_{k_0}\text{ if }i=i_0\\\ \ast\text{ otherwise}\end{cases}$$ $$f_i(p_k)=\ast\hskip0.3in f_i(\ast)=\ast$$ Thus $$f_i^{-1}(p_k)=\coprod_{j\in J}P_{i,j,k}\subset X,$$ and thus $f_i^{-1}(p_k)\cap \Omega_j=P_{i,j,k}$, so $|f_i^{-1}(p_k)\cap\Omega_j|=\gamma(i,j,k)$.

Let $P_{i,j,k}=\{1_{i,j,k},\ldots,n_{i,j,k},\ldots,\gamma(i,j,k)_{i,j,k}\}$ (so that for each choice of $i$, $j$, and $k$, we have a disjoint set of size $\gamma(i,j,k)$). Let $Q=\{p_k\mid k\in K\}$ (so just $K$, up to relabeling). Let $R=\{\ast\}$. Let $$X=Q\coprod R\coprod_{\substack{i\in I,j\in J\\\k\in K}}P_{i,j,k}.$$ Define $$\Omega_j=\coprod_{i\in I,k\in K}P_{i,j,k}\subset X,$$ and $$f_{i}(n_{i_0,j_0,k_0})=\begin{cases}p_{k_0}\text{ if }i=i_0\\\ \ast\text{ otherwise}\end{cases}$$ $$f_i(p_k)=\ast\hskip0.3in f_i(\ast)=\ast$$ Thus $$f_i^{-1}(p_k)=\coprod_{j\in J}P_{i,j,k}\subset X,$$ and thus $f_i^{-1}(p_k)\cap \Omega_j=P_{i,j,k}$, so $|f_i^{-1}(p_k)\cap\Omega_j|=\gamma(i,j,k)$.