Not all of the outcomes (0,0), (0,1), (0,2),... (3,2), (3,3) are equally likely. Although there are indeed sixteen possible outcomes, unless all outcomes are equiprobable you may not simply do the fraction of favorable outcomes compared to total outcomes regardless. For example, the outcome $(0,1)$ is three times as likely to occur as the outcome $(0,0)$.
Approach via a combination of the binomial theorem and the multiplication and addition principles of probability.
If $B$ has zero heads, $A$ wins with at least one head. The probability of $B$ having zero heads is $\frac{1}{8}$. The probability of $A$ having at least one head is $\frac{7}{8}$. The probability of $B$ having zero heads and $A$ having at least one head is then $\frac{7}{64}$
If $B$ has one head, $A$ wins with at least two heads. The probability of $B$ having one head is $\frac{3}{8}$. The probability of $A$ having at least two heads is $\frac{4}{8}$. The probability then of both occurring is $\frac{12}{64}$
If $B$ has two heads, $A$ wins only with getting three heads. The probability of $B$ having two heads is $\frac{3}{8}$. The probability of $A$ having three heads is $\frac{1}{8}$. The probability then of both occurring is $\frac{3}{64}$
If $B$ has three heads, $A$ cannot win.
As the sample space can be partitioned into the result of how many heads $B$ received as above, the probability that $A$ wins is the sum of the aforementioned probabilities:
$Pr(\text{A wins}) = \frac{7}{64}+\frac{12}{64}+\frac{3}{64}+0 = \frac{22}{64}=\frac{11}{32}$
(didn't see John's answer while typing this extra paragraph, the method is essentially the same)
An alternate way to view the solution: $1 = Pr(\text{A wins})+Pr(\text{A and B tie}) + Pr(\text{B wins})$
Due to the symmetry of their circumstances, we know that $Pr(\text{A wins})=Pr(\text{B wins})$. This leaves us with the question of finding $Pr(\text{A and B tie})$
This can occur with either zero heads each, one head each, two heads each, or three heads each.
These occur with probabilities $\frac{1}{64}, \frac{9}{64},\frac{9}{64},\frac{1}{64}$ respectively.
The probability that $A$ and $B$ tie is then $\frac{20}{64}$
This tells us that $Pr(A~\text{wins}) = \frac{1}{2}(1-\frac{20}{64})=\frac{22}{64}=\frac{11}{32}$