A boolean function is simply a function on boolean variables that produces a boolean result. There is no requirement that it must built in a certain way.

However, if you insist that you want some chain of logical operators, for any fixed $m,k$ we can build such an $f$. Suppose for now that $m ge k$.

Note that $x$ has at least as many active values as $y$, if and only if there is a way to match up the elements of $x$ with $y$ so that every active element of $y$ is matched with an active element of $x$. That is, if $x_i$ and $y_j$ are matched, then either $y_j = 0$ or $x_i = 1$, or in symbols, $y_j implies x_i$.

To formalize this, define a pairing to be a set $P subset {1, ldots, m} times {1, ldots k}$ such if $(i, j), (r,s) in P$, then $i = r iff j = s$, and for all $1 le j le k$, there exists some $i$ such that $(i,j) in P$. Let $scr P$ be the set of all such pairings. If $p in P in mathscr P$, denote its coordinates by $p = (i_p, j_p)$.

So $x$ has at least as many active elements as $y$ if and only if there exists a $Pin scr P$ such that $$bigwedge_{pin P} (y_{j_p} implies x_{i_p})$$ That is, when $$bigvee_{P in mathscr P}bigwedge_{pin P} (y_{j_p} implies x_{i_p})$$

is true. So $$f(x,y) = bigvee_{P in mathscr P}bigwedge_{pin P} (y_{j_p} implies x_{i_p})$$

When $m < k$, there are no pairings. Instead define a partial pairing as a tuple $(P, Q)$ where $P$ is a pairing of some $m$ elements of ${1,ldots, k}$ with ${0,ldots, m}$, and $Q$ is the set of $k-m$ indices that were not paired. Let $scr Q$ be the set of partial pairings.

Now we can express $$f(x,y) = bigvee_{(P,Q) in mathscr Q}left(bigwedge_{pin P} (y_{j_p} implies x_{i_p})wedge bigwedge_{j in Q} (lnot y_j)right)$$

<Edit> The above only encodes $x$ having as many active values as $y$. To also encode that at least one value must be active, we have

$$f(x,y) = left(bigvee_{i=1}^m x_iright) wedge left(bigvee_{(P,Q) in mathscr Q}left(bigwedge_{pin P} (y_{j_p} implies x_{i_p})wedge bigwedge_{j in Q} (lnot y_j)right)right)$$ </Edit>

All this bit about pairings is just a way of expressing the short-hand notation that holds for every $m,k$. For a particular $m,n$, it is just a matter of listing every every possible pairing. For example, when $m = k = 2$, this is

$$f(x, y) = left(x_1 vee x_2right) wedge left(left[ (y_1 implies x_1) wedge (y_2 implies x_2) right] vee left[(y_1 implies x_2) wedge (y_2 implies x_1) right]right)$$

but as $m, k$ get larger, this quickly grows into a ridiculously long expression. But despite that, it still accurately represents $f(x,y)$.

There are undoubtedly simpler expressions that are equivalent. But what this shows is that it is always possible to express $f(x,y)$ in terms of logical operators. Which is something that is true of any boolean function.

<Edit> Adding a few simple examples.

To simplify the functions, I'm just doing "$x$ has at least as many active elements as $y$". To get the actual condition of the question, "and" the functions with $(x_1 vee x_2 vee ldots vee x_m)$.

If we match a set $a$ of three variables with set $b$ of two, there are $6$ possible pairings between them, each with one member of $a$ left over: $$begin{array}{c|ccc} & a_1 & a_2 & a_3\ hline p_1 & b_1 & b_2 & - \ p_2 & b_2 & b_1 & - \ p_3 & b_1 & - & b_2 \ p_4 & b_2 & - & b_1 \ p_5 & - & b_1 & b_2 \ p_6 & - & b_2 & b_1 end{array}$$

If $a = x$ and $b = y$, then each pairing is represented by the expressions $$begin{array}{c|c} p_1 & (y_1 implies x_1) wedge (y_2 implies x_2) \ p_2 & (y_2 implies x_1) wedge (y_1 implies x_2) \ p_3 & (y_1 implies x_1) wedge (y_2 implies x_3) \ p_4 & (y_2 implies x_1) wedge (y_1 implies x_3) \ p_5 & (y_1 implies x_2) wedge (y_2 implies x_3) \ p_6 & (y_2 implies x_2) wedge (y_1 implies x_3) end{array}$$ When $x$ has the extra elements, it doesn't matter if they are active or not, so nothing has to be added for them. When $x$ has more or equal active elements, at least one of these statements will be true, and vice versa. So the total condition is

$$begin{align}f(x,y) = &[(y_1 implies x_1) wedge (y_2 implies x_2)] vee \ &[(y_2 implies x_1) wedge (y_1 implies x_2)] vee \ &[(y_1 implies x_1) wedge (y_2 implies x_3)] vee \ &[(y_2 implies x_1) wedge (y_1 implies x_3)] vee \ &[(y_1 implies x_2) wedge (y_2 implies x_3)] vee \ &[(y_2 implies x_2) wedge (y_1 implies x_3)]end{align}$$

If $a = y, b = x$, so it is $y$ that has the unmatched elements, then it is necessary to guarantee that the unmatched elements are not active:

$$begin{array}{c|c} p_1 & (y_1 implies x_1) wedge (y_2 implies x_2) wedge lnot y_3 \ p_2 & (y_1 implies x_2) wedge (y_2 implies x_1) wedge lnot y_3 \ p_3 & (y_1 implies x_1) wedge lnot y_2 wedge (y_3 implies x_2) \ p_4 & (y_1 implies x_2) wedge lnot y_2 wedge (y_3 implies x_1) \ p_5 & lnot y_1 wedge (y_2 implies x_1) wedge (y_3 implies x_2) \ p_6 & lnot y_1 wedge (y_2 implies x_2) wedge (y_3 implies x_1) end{array}$$

And again, we "or" the various pairings together to get the full function: $$begin{align}f(x,y) = &[(y_1 implies x_1) wedge (y_2 implies x_2) wedge lnot y_3] vee \ &[(y_1 implies x_2) wedge (y_2 implies x_1) wedge lnot y_3] vee \ &[(y_1 implies x_1) wedge lnot y_2 wedge (y_3 implies x_2)] vee \ &[(y_1 implies x_2) wedge lnot y_2 wedge (y_3 implies x_1)] vee \ &[lnot y_1 wedge (y_2 implies x_1) wedge (y_3 implies x_2)] vee \ &[lnot y_1 wedge (y_2 implies x_2) wedge (y_3 implies x_1)]end{align}$$ <Edit>