Suppose I have a (finitely-presented, say) graded module $M$ over $k[x,y]$, and I happen to know the rank $R\_{(a,b),(c,d)}$ of each map $x^{c−a}y^{d−b}:M\_{a,b}→M\_{c,d}$ for each pair of integers with $a\leq c$ and $b \leq d$, as well as the Hilbert function $\dim\_k(M\_{a,b})$. 


Recall that the Betti numbers of $M$ are defined to be the grades and multiplicities of the generators of a minimal free resolution.


Is there an explicit formula which expresses these Betti numbers directly in terms of the ranks $R\_{(a,b),(c,d)}$ and the Hilbert function? 


As motivation, there is a well-known expression for the alternating sums of the Betti numbers in terms of the Hilbert function, e.g. Corollay 1.10 in "The Geometry of Syzygies" by Eisenbud. 



Combinatorial formula for Betti numbers of a $k[x,y]$-module
It seems unlikely that the Betti numbers can be determined by the data you list (bigraded Hilbert function and ranks of multiplication maps). Consider the following possibility: $\dim M\_{1,0} = \dim M\_{0,1} = 1$, $\dim M\_{1,1} = 2$, and all other $M\_{a,b}$ are zero; and the maps $y : M\_{1,0} \to M\_{1,1}$, $x : M\_{0,1} \to M\_{1,1}$ each have rank $1$ (all other multiplication maps are apparently zero).


Given this data and nothing else, it's impossible to tell if the maps from $M\_{1,0}$ and $M\_{0,1}$ to $M\_{1,1}$ have the same image in $M\_{1,1}$, or not.


An example where they have different images is $$M = (x)/(x^2,xy^2) \oplus (y)/(x^2y,y^2)$$ (where parentheses denote ideals). An example where they have the same image is $$M' = (x,y)/(x^2,y^2) \oplus (xy)/(x^2y,xy^2).$$


$M$ is generated by $x,y$ in bidegrees $(1,0)$ and $(0,1)$, while $M'$ requires generators $x,y$, and an "extra" copy of $xy$, in bidegrees $(1,0)$, $(0,1)$, and $(1,1)$. So, not even the first Betti degrees are determined by the data that you have.


Sometimes you’ll be able to read off some information. If one of the incoming maps is zero or surjective then you’ll know how many generators there are in that bidegree. At the moment I can’t think of any other cases, or anything for higher Betti numbers. 


