Something's wrong in the last formula of the question. The elements of $f$ are functions $N\to U$, but the elements of $\prod_{i\in N}(\mathcal L_{c(i)}(i))$ are functions on $N$ whose values are subsets of $U$, elements of the filters $\mathcal L_{c(i)}(i)$, not elements of $U$. I'm going to assume that you meant that for every choice of sets $Y_i\in\mathcal L_{c(i)}(i)$ (for all $i\in N$), $(\prod_{i\in N}Y_i)\cap f\neq\emptyset$.

If that's the intended question, then here's a counterexample. Let $U$ and $N$ both be the set of natural numbers. Let $\mathcal L_0(0)=\mathcal L_1(0)=$ the filter of cofinite subsets of $U$. For all $n>0$, let $\mathcal L_0(n)$ be the principal filter generated by $\{0\}$ and let $\mathcal L_1(n)$ be the principal filter generated by $\{1\}$. Let $f$ be the family of those functions $x:N\to U$ such that $x(n)=0$ whenever $1\leq n\leq x(0)$, and $x(n)=1$ whenever $n>x(0)$. (The only important thing here is that $x(0)$ determines the values of $x$ at all points other than 0 and those values are 0 or 1.)

If $X\in\text{up }\mathcal L_0\cap\text{up }\mathcal L_1$, then $X_0$ contains all but finitely many elements of $\mathbb N$ while $X_k$ for any $k>0$ contains both 0 and 1. It follows immediately that $\prod_{i\in N}X_i$ contains an element of $f$, in fact infinitely many elements of $f$.

Now consider any fixed $c\in\{0,1\}^n$, and notice that there is at most one $k\in N$ such that the sequence $x=(k,c(1),c(2),\dots)$ (i.e., $c$ with $c(0)$ replaced by $k$) is in $f$. Let $Q=U-\{k\}$ if there is such a $k$ and $Q=U$ otherwise. Then one possibility for the $Y_i$ in (my interpretation of) the question is $Y_0=Q$ and $Y_i=\{c(i)\}$ for all $i>0$. Then $\prod_{i\in N}Y_i$ does not contain any element of $f$.