Given that $f(k)$ is a multiple of 3 whenever $k$ is a multiple of 4, we need to consider how to map elements from the domain {1, 2, 3, ..., 20} to the codomain {1, 2, 3, ..., 20} following this rule.

1. We first consider the subset of the domain that consists of multiples of 4: {4, 8, 12, 16, 20}. There are 5 elements in this subset.

2. We then consider the subset of the codomain that consists of multiples of 3: {3, 6, 9, 12, 15, 18}. There are 6 elements in this subset.

3. According to the given condition, each of the 5 multiples of 4 must be mapped to a multiple of 3. This can be done in ${ }^6C_5 \cdot 5! = 6!$ ways, considering that there are 6 options for each of the 5 multiples of 4 (each choice constitutes a combination), and we then consider the permutations of these 5 choices.

4. The remaining 15 elements in the domain (20 original elements minus the 5 multiples of 4) can be mapped onto the remaining 15 elements in the codomain (20 original elements minus the 6 multiples of 3, plus one multiple of 3 that has been assigned to a multiple of 4). This can be done in $15!$ ways.

So, combining these two cases, the total number of onto functions $f$ is $6! \times 15!$, which corresponds to option C.