Assumptions are: One team wins, maximum of 5 kicks for each team, team A and B, team A do the first kick.
My Try:
Case 1- Team A wins, further sub-cases as all 10 kicks are made, only 9, only 8, only 7, only 6.
If all 10 kicks are made, that means Team B failed to score in the final kick and Team B leads by 1 goal after 9 kicks. So the possibilities are 5-4,4-3,3-2,1-0 and in each of the possibility we have to find number of ways the goals are distributed among the 5 and 4 chances of Team A and B respectively.
Next sub-case is only 9 kicks are made, again find all possibility and so on
Same for case 2.
Is there any better way to solve it?
Let's say that team $A$ gets a point whenever they make a kick, and they also get a point whenever team $B$ misses a kick. The kickoff ends as soon as $A$ gets six points. This is because if $A$ has made $n$ kicks and Bob has missed $6-n$ kicks, then $B$ has only $n-1$ kicks left, which is not enough to make a comeback.
There are $10$ opportunities for $A$ to score a point, so there are $\binom{10}6$ ways to choose the ways for $A$ to score six points. Similarly, there are $\binom{10}6$ ways for Bob to win, so the number of possible tiebreakers without a tie is $$ 2\cdot \binom{10}6 $$