A reasonable sample space is the set of all ordered pairs $(A,B)$, where $A$, the first set chosen, is a subset of $Q$ with $k$ elements, as is $B$, the second set chosen. This sample space has size $\binom{100}{k}\binom{100}{k}$ elements, and they are all equally likely. For easy comparison with the answer by JohnJamesSmith, we use his notation.
So given $i$, we want to find $P(W=i)$. Clearly $P(W=i)=0$ if $i. Also, if $i>\min(2k,100)$, then $P(W=i)=0$. We need to deal with the remaining cases.
In how many ways can the cardinality of $A\cup B$ be $i$? Note that $A$ can be chosen in any one of $\binom{100}{k}$ ways. In order for $A\cup B$ to have cardinality $i$, the set $B$ must contain $i-k$ objects that are not in $A$, and the region of overlap between $A$ and $B$ must have cardinality $2k-i$.
Given $A$, we can choose $2k-i$ elements of $A$ in $\binom{k}{2k-i}$ ways. For every one of these ways, the elements of $B$ outside $A$ can be chosen in $\binom{100-k}{i-k}$ ways. So the number of elements in the sample space that give $W=i$ is equal to $\binom{100}{k}\binom{k}{2k-i}\binom{100-k}{i-k}$ ways. To find $P(W=i)$, divide. Note the cancellation of $\binom{100}{k}$. Thus $P(W=i)=\frac{\binom{k}{2k-i}\binom{100-k}{i-k}}{\binom{100}{k}}.$ We needn't have bothered with $A$, it was just introduced to provide a respectable-sounding sample space. For just calculating the probability, we can without loss of generality assume that $A$ consists of the first $k$ elements of $Q$.