I think that what you are asking for is impossible. Given any element of $\prod_n X_n$ the element is uniquely determined by its image in each of the individual $X_n$'s. So if two elements of $Y$ agree on each $X_n$ then they must be the same element.
In language similar to yours, what you probably want for the finite case is "If $X$ and $Y$ are finite sets such that $|X| < |Y|$ and $f:Y\rightarrow X$ is any map then there exists an element $x\in X$ such that $|f^{-1}(x)| > x$." More generally, given any finite sequence $X_1,\ldots,X_n$ of finite sets and any set $Y$ such that $|Y| > |X_1|\cdot|X_2|\cdots|X_n|$ and any sequence of maps $f_i:Y\rightarrow X_i$ then there exists a sequence of elements $x_1,\ldots,x_n$ and two elements $y,y'\in Y$ such that $f_i(y) = f_i(y')$ for any $i$.
The problem with the infinite case is that there are injective but not surjective maps between infinite sets with the same cardinality. However, it is true that given a sequence of finite sets $X_1,X_2,\ldots$ and an uncountable a set $Y$, Y$ with cardinality greater than that of $\prod X_n$, if you have any sequence of maps $f_i:Y\rightarrow X_i$ then there exists an uncountable subset $Z\subseteq Y$ such that for any two elements $z,z'$ of $Z$ you have $f_i(z) = f_i(z')$ for all $i$.
In even more generality, I believe that if you have any set of sets ${X_\alpha}$ and any set $Y$ such that the cardinality of $Y$ is larger than the cardinality of $\prod_\alpha X_\alpha$ then you have a similar statement.
