First of all, worst case would be 3 comparisons. For example, A = [10,20], B = [15,30]. Here number of comparisons would be 3.
Now for expected number of comparisons, suppose we are given two lists A = [x1,x2], B = [x3,x4]. For simplicity, suppose all 4 elements are distinct.
If we were to write all possible results of merging x1,x2,x3,x4, we have following $\binom{4}{2}=6$ possibilities (Note that in each possibility, x1 has to come before x2, and x3 before x4 because A and B are sorted)
x1,x2,x3,x4
x1,x3,x2,x4
x1,x3,x4,x2
x3,x4,x1,x2
x3,x1,x4,x2
x3,x1,x2,x4
Now we can have either 2 comparisons or 3 comparisons. We have 2 comparisons if either both x1,x2 < x3 or both x3,x4<x1, so P(2 comparisons) = 2/6
We will have 3 comparisons in rest of the 4 cases (you can verify), so P(3 comparisons) = 4/6
So $\text{E[num comparisons]} = 2*\text{P(2 comparisons)} + 3*\text{P(3 comparisons)}$
$= 2*\frac{2}{6} + 3*\frac{4}{6} = \frac{16}{6} = \frac{8}{3}$
We can generalize it for two lists $A$ and $B$ of size $m$. Number of ways in which final merged list $C$ can be formed is $\binom{2m}{m}$, because we have total $2m$ locations in $C$, out of which we can choose any $m$ locations to put elements of $A$, rest are filled by $B$.
Now we count number of comparisons done during merging. Suppose during merging, $A$ finishes first (other case is symmetrical, and we will multiply the count by 2). So number of comparisons will be size of list $C$ when $A$ finishes. So we argue on size of $C$ (let we call it $S$ when $A$ finishes).
Minimum value of $S$ is $m$ because when $A$ finishes, $C$ must contain at least $m$ elements, and maximum value of $S$ is $2m-1$, which happens when $B$ is left with just 1 element when $A$ finishes. So we make cases on $S$ :
$S = m$ : Only 1 case, when all elements of $A$ are smaller than of B.
$S = m+1$ : $C$ contains $m+1$ elements, out of which $m$ are of A, $1$ is of B, that 1 element of $B$ can take any location out of $m$ locations in C, so $\binom{m}{1}$ choices.
$S = m+2$ : $C$ contains $m+2$ elements, out of which $m$ are of A, $2$ are of B, those $2$ elements of $B$ can take any location out of $m+1$ locations in C, so $\binom{m+1}{2}$ choices.
This can go on till $S=2m-1$ : $\binom{2m-2}{m-1}$ choices.
So total number of comparisons in all cases
$=2*\left(m*1 + (m+1)*\binom{m}{1} + \ldots + (2m-1)*\binom{2m-2}{m-1}\right)$
Multiplication by $2$ outside is due to symmetry.
Finally, expected value of number of comparisons
$=\frac{2*\left(m*1 + (m+1)*\binom{m}{1} + \ldots + (2m-1)*\binom{2m-2}{m-1}\right)}{\binom{2m}{m}}$
We can generalize it similarly for different size lists.