Let $g = \gcd(45, n)$, then the answer will be $9n/g$.

$$\begin{align*} V &= (2, 7) \quad \text{i.e. hopping from } (a, b) \text{ to } (a + 2, b + 7) \\ V^{-1} &= (-2, -7) \quad \text{i.e. hopping from } (a, b) \text{ to } (a - 2, b - 7) \\ H &= (7, 2) \quad \text{i.e. hopping from } (a, b) \text{ to } (a + 7, b + 2) \\ H^{-1} &= (-7, -2) \quad \text{i.e. hopping from } (a, b) \text{ to } (a - 7, b - 2), \end{align*}$$
where the second coordinate is to be understood modulo $n$. We use these symbols to denote sequences of moves. For example, $V^2H^{-3}V$ means $V$ followed by three consecutive $H^{-1}$, followed by two consecutive $V$. Note that we read this from right to left.

Lower Bound. We first show that Jimmy can reach at least $9n/g$ squares. Starting at the corner square $(1, 1)$, the sequence $S = H^{-1}V^3H^{-1}V^4$ lets Jimmy travel as follows: $(1, 1) \mapsto (3, 8) \mapsto (5, 15) \mapsto (7, 22) \mapsto (9, 29) \mapsto (2, 27) \mapsto (4, 34) \mapsto (6, 41) \mapsto (8, 48) \mapsto (1, 46)$. This means that $S$ corresponds to the addition of $(0, 45)$. Hence, iterating $S$ we see that $S^k$ corresponds to the addition of $45k \pmod n$ to the column number and no change in the row number.
The congruence $b \equiv 1 + 45k \pmod{n}$, has a solution $k$ if and only if $g \mid b - 1$. We can always pick a positive solution $k$ in its congruence class modulo $n$. Therefore, Jimmy can reach $n/g$ squares in row 1, namely those in columns $b = 1 + mg, 0 \le m < n/g$.
Through the sequence $S$, Jimmy has visited each row exactly once before coming back to row 1. Considering the part of the sequence $S^2$ that starts and ends at row $a$, the same argument as above shows that Jimmy is able to hop on $n/g$ squares on row $a$ as well. Hence, Jimmy can reach at least $9n/g$ squares in total.

Suppose Jimmy starts at square $(a, b)$ with move $V^{p_1}H^{q_1}V^{p_2}H^{q_2}\cdots V^{p_r}H^{q_r}$, where $p_i, q_i$ are integers. Let $p = p_1 + p_2 + \cdots + p_r$ and $q = q_1 + q_2 + \cdots + q_r$. Then Jimmy ends up on the square $(a + 2p + 7q, b + 7p + 2q)$, where the second coordinate is to be taken modulo $n$. If Jimmy ends up in row $a$, we have $2p + 7q = 0$, hence $2 \mid q$ and $7 \mid p$ and there exists an integer $k$ such that $p = 7k, q = -2k$. The column number is then congruent to
$$b + 7p + 2q \equiv b + 7 \cdot 7 \cdot k - 2 \cdot 2 \cdot k \equiv b + 45k \pmod{n}.$$
Hence, Jimmy can only move from square $(a, b)$ to square $(a, c)$ if $c - b \equiv 45k \pmod{n}$ for some integer $k$. Such a $k$ exists exactly when $g \mid c - b$. As any two squares Jimmy reaches during his entire journey are connected by a sequence of moves, it follows that Jimmy can reach at most $n/g$ squares in each row. Hence, the maximum number of squares Jimmy can reach is $9n/g$.