Let's say:
$n_1$: number of successful passes by Italy in first half
$n_2$: number of attempted passes by Italy in first half
$n_3$: number of successful passes by U.S in first half
$n_4$: number of attempted passes by U.S in first half
$n_5$: number of successful passes by Italy in second half
$n_6$: number of attempted passes by Italy in second half
$n_7$: number of successful passes by U.S in second half
$n_8$: number of attempted passes by U.S in second half
Of course, we must have:
$$n_1 \leq n_2$$
$$n_3 \leq n_4$$
$$n_5 \leq n_6$$
$$n_7 \leq n_8$$
Now, we want Italy to have a higher rate or successful passes in each of the halves, but the U.S. have a better rate over the whole game:
$$\frac{n_1}{n_2} > \frac{n_3}{n_4}$$
$$\frac{n_5}{n_6} > \frac{n_7}{n_8}$$
$$\frac{n_3+n_7}{n_4+n_8} > \frac{n_1+n_5}{n_2+n_6}$$
which means:
$$n_1 \cdot n_4 > n_3 \cdot n_2$$
$$n_5 \cdot n_8 > n_7 \cdot n_6$$
$$(n_3 + n_7) \cdot (n_2+ n_6) > (n_1 +n_5)\cdot (n_4 +n_8)$$
Working out the latter inequality, and then using the first two, this implies:
$$n_3\cdot n_2 + n_3 \cdot n_6 + n_7 \cdot n_2 + n_7 \cdot n_6$$
$$ > n_1\cdot n_4 + n_1 \cdot n_8 + n_5 \cdot n_4 + n_5 \cdot n_8$$
$$ > n_3\cdot n_2 + n_1 \cdot n_8 + n_5 \cdot n_4 + n_7 \cdot n_6$$
And, so we must have:
$$n_3 \cdot n_6 + n_7 \cdot n_2 > n_1 \cdot n_8 + n_5 \cdot n_4$$
However, note that this latter inequality cannot replace the earlier equality; it's merely something that follows from the earlier inequalities. In other words, the solution set is still any $n_1$ through $n_8$ for which we have:
$$n_1 \leq n_2$$
$$n_3 \leq n_4$$
$$n_5 \leq n_6$$
$$n_7 \leq n_8$$
$$n_1 \cdot n_4 > n_3 \cdot n_2$$
$$n_5 \cdot n_8 > n_7 \cdot n_6$$
$$(n_3 + n_7) \cdot (n_2+ n_6) > (n_1 +n_5)\cdot (n_4 +n_8)$$
But we can use as a further 'check' that:
$$n_3 \cdot n_6 + n_7 \cdot n_2 > n_1 \cdot n_8 + n_5 \cdot n_4$$
That is, any values of $n_1$ through $n_8$ that do not obey the latter inequality will not be a solution.
I'd like to note that the inequalities cannot be satisfied if both teams made the same number of attempts in each of the halves. That is, we cannot have that $n_2=n_4=n_I$ and $n_6=n_8=n_{II}$, for then we would need to have:
$$n_1 > n_3$$
$$n_5 > n_7$$
and that would contradict our 'check':
$$n_3 \cdot n_{II} + n_7 \cdot n_I > n_1 \cdot n_{II} + n_5 \cdot n_I$$
However, we can have the same number of attempts for one of the halves. As a simple example: Italy completed $1$ out of $2$ passes in the first half, but the U.S. $0$ out of $2$, and Italy completed $1$ out of $1$ passes in the second half, and the U.S. $7$ out of $8$.
This latter example also shows that Italy can have a $100$% completion rate in one half, and the U.S. a $0$% completion rate in one of its halves, and yet the U.S. still obtaining a better completion rate for the whole game.
Finally, the 'smallest' numbers satisfying the inequalities (where by smallest, I mean minimizing the maximum of $n_1$ through $n_8$) is with Italy completing $1$ out of $3$ passes in the first half, and the U.S. $0$ out of $1$, while Italy completed $1$ out of $1$ passes in the second half, and the U.S. $3$ out of $4$.