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$then you have a similar statement. 1 I think that what you are asking for is impossible. Given any element of$\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$and an uncountable set$Y$, if you have any sequence of maps$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.