Let $B = A + C$ where $\frac{A}{n + 1} \leq C < \frac{A}{n}$.

First, here is an informal sketch of the proof.

Lavaman's strategy: Use only safe intervals with $nA + C - 1$ integers. The flea will start at position $[1, C]$ from the left, which puts him at position $[nA, nA + C - 1]$ from the right. After $n-1$ jumps, he will still have $nA - (n-1)(A+C) = A - (n-1)C > C$ distance to go, which is not enough for a big jump to clear the lava. Thus, he must do at least $n$ jumps in the safe interval, but that's possible only with all small jumps, and furthermore is impossible if the starting position is $C$. This gives him starting position 1 higher in the next safe interval, so sooner or later the flea is going to hit the lava.

Flea's strategy: The flea just does one interval at a time. If the safe interval has at least $nA + C$ integers in it, the flea has distance $d > nA$ to go to the next lava when it starts. Repeatedly do big jumps until $d$ is between $1$ and $C \bmod A$, then small jumps until the remaining distance is between $1$ and $C$, then a final big jump. This works as long as the first part does. However, we get at least $n$ big jumps since $\lfloor(d-1)/A\rfloor$ can never go down two from a big jump (or we'd be done doing big jumps), so we get $n$ big jumps, and thus we are good if $d \bmod A$ is in any of $[1, C], [C+1, 2C], \dots, [nC+1, (n+1)C]$, but that's everything. $\square$

Let $C = B - A$. We shall write our intervals of lava in the form $(L_i, R_i] = \{L_i + 1, L_i + 2, \dots, R_i\}$, where $R_i = L_i + A$ and $R_{i-1} < L_i$ for every $i \geq 1$. We also let $R_0 = 0$. We shall also represent a path for the flea as a sequence of integers $x_0, x_1, x_2, \dots$ where $x_0 = 0$ and $x_j - x_{j-1} \in \{A, B\}$ for every $j \geq 0$.

Now here is a detailed proof.

First, assume $F < (n-1)A + B (= nA + C)$: we must prove that Lavaman has a winning strategy. Let $L_i = R_{i-1} + nA + C - 1$ for every $i \geq 1$. (Observe that $nA + C - 1 \geq F$.)

means that $x_j \notin (L_i, R_i]$ for all $i, j \geq 1$. For each $i \geq 1$, let
$$M_i = \max\{x_j : x_j \leq L_i\}, \quad m_i = \min\{x_j : x_j > R_i\}, \ \text{and} \quad J(i) = \max\{j : x_j \leq L_i\}.$$
Also let $m_0 = 0$. Then for $i \geq 1$ we have
$$M_i = x_{J(i)} \quad \text{and} \quad m_i = x_{J(i)+1}.$$
Also, for every $i \geq 1$, we have
(a) $m_i = M_i + B$ (because $M_i + A \leq L_i + A = R_i$);
(b) $L_i \geq M_i > L_i - C$ (since $M_i = m_i - B > R_i - B = L_i + A - B$);
(c) $R_i < m_i \leq R_i + C$ (since $m_i = M_i + B \leq L_i + B = R_i + C$).

$$\begin{align*} M_{i+1} = x_{J(i+1)} = x_{J(i)+n+1} &\geq x_{J(i)+1} + (n-1)A + B \\ &> R_i + nA + C \\ &= L_{i+1} + 1 \\ &> M_{i+1} \end{align*}$$
which is a contradiction. This proves Claim 2.

$$\begin{align*} x_{J(i+1)+1} &= x_{J(i)+n+2} = x_{J(i)+1} + nA + B; \\ \text{i.e., } m_{i+1} &= m_i + nA + B \quad \text{for each } i \geq 1. \end{align*}$$
Therefore
$$\begin{align*} m_{i+1} - R_{i+1} &= m_i + nA + B - (R_i + nA + C - 1 + A) \\ &= m_i - R_i + 1. \end{align*}$$
Hence
$$C \geq m_{C+1} - R_{C+1} = m_1 - R_1 + C > C$$
which is a contradiction. Therefore no path for the flea avoids all the lava. We observe that Lavaman only needs to put lava on the first $C + 1$ intervals.

Now assume $F \geq (n-1)A + B$. We will show that the flea can avoid all the lava. We shall need the following result:

Claim 3: Let $d \geq nA$. Then there exist nonnegative integers $s$ and $t$ such that $sA + tB \in (d - C, d]$.

We shall prove this result at the end.

First, observe that $L_1 \geq nA$. By Claim 3, it is possible for the flea to make a sequence of jumps starting from $0$ and ending at a point of $(L_1 - C, L_1]$. From any point of this interval, a single jump of size $B$ takes the flea over $(L_1, R_1]$ to a point in $(R_1, R_1+C]$, which corresponds to the point $x_{J(1)+1} (= m_1)$ on the flea's path.

Now we use induction to prove that, for every $i \geq 1$, there is a path such that $x_j$ avoids lava for all $j \leq J(i)+1$. The case $i=1$ is done, so
assume that the assertion holds for a given $i$. Then $x_{J(i)+1} = m_i \in (R_i, R_i + C]$. Therefore
$$L_{i+1} - m_i \geq R_i + F - (R_i + C) = F - C \geq nA.$$
Applying Claim 3 with $d = L_{i+1} - m_i$ shows that the flea can jump from $m_i$ to a point of $(L_{i+1} - C, L_{i+1}]$. A single jump of size $B$ then takes the flea to a point of $(R_{i+1}, R_{i+1} + C]$ (without visiting $(L_{i+1}, R_{i+1}]$), and this point serves as $x_{J(i+1)+1}$. This completes the induction.

Proof of Claim 3: Let $u$ be the greatest integer that is less than or equal to $d/A$. Then $u \geq n$ and $uA \leq d < (u+1)A$. For $v = 0, \dots, n$, let
$$z_v = (u - v)A + vB = uA + vC.$$
Then
$$z_0 = uA \leq d,$$
$$z_n = uA + nC = uA + (n + 1)C - C \geq (u + 1)A - C > d - C.$$
and $z_{v+1} - z_v = C$ for $v = 0, \dots, n-1$.

Therefore we must have $z_v \in (d-C, d]$ for some $v$ in $\{0, 1, \dots, n\}$. $\square$