Let $a_1, a_2, a_3, a_4$, and $a_5$ represent the integers at vertices $v_1$ to $v_5$ (in order around the pentagon) at the start of the game. We will first show that the game can be won at only one of the vertices. Observe that the quantity $a_1 + 2a_2 + 3a_3 + 4a_4 \pmod{5}$ is an invariant of the game. For instance, one move involves replacing $a_1, a_3$ and $a_5$ by $a_1 - m, a_3 + 2m$ and $a_5 - m$. Thus the quantity $a_1 + 2a_2 + 3a_3 + 4a_4$ becomes
$$(a_1 - m) + 2a_2 + 3(a_3 + 2m) + 4a_4 = a_1 + 2a_2 + 3a_3 + 4a_4 + 5m,$$
which is unchanged modulo $5$. The other moves may be checked similarly. Now suppose that the game may be won at vertex $v_j$. The value of the invariant at the winning position is $2011j$. If the initial value of the invariant is $n$, then we must have $2011j \equiv n \pmod{5}$, or $j \equiv n \pmod{5}$. Hence the game may only be won at vertex $v_j$, where $j$ is the least positive residue of $n \pmod{5}$.

By renumbering the vertices, we may assume without loss of generality that the potentially winning vertex is $v_5$. We will show that the game can be won in four moves by adding a suitable amount $2m_j$ at vertex $v_j$ (and subtracting $m_j$ from the opposite vertices) on the $j$th turn for $j = 1, 2, 3, 4$. The net change at vertex $v_1$ after these four moves is $2m_1 - m_3 - m_4$, which must equal $-a_1$ if we are to finish with $0$ at $v_1$. In this fashion we obtain the system of equations
$$\begin{aligned} 2m_1 - m_3 - m_4 &= -a_1 \\ 2m_2 - m_4 &= -a_2 \\ 2m_3 - m_1 &= -a_3 \\ 2m_4 - m_1 - m_2 &= -a_4 \\ -m_2 - m_3 &= -a_5 + 2011, \end{aligned}$$
which has an integral solution if and only if the game may be won. The sum of the first four equations is the negative of the fifth equation, so the fifth equation is redundant. Multiplying the first four equations by $-1, 3, -3, 1$ and adding them yields $5m_2 - 5m_3 = a_1 - 3a_2 + 3a_3 - a_4$. But we are assuming $v_5$ is the potentially winning vertex, so we see
$$a_1 - 3a_2 + 3a_3 - a_4 \equiv a_1 + 2a_2 + 3a_3 + 4a_4 \equiv n \equiv 5 \equiv 0 \pmod{5}.$$
Therefore we may divide by $5$ to obtain $m_2 - m_3 = \frac{1}{5}(a_1 - 3a_2 + 3a_3 - a_4)$. We also know that $m_2 + m_3 = a_1 + a_2 + a_3 + a_4$, and one easily confirms that the right-hand sides of these equations are integers with the same parity. Hence the system admits a solution with $m_2$ and $m_3$ integral. The second and third equations then quickly give integer values for $m_1$ and $m_4$ as well, so the system has an integral solution, meaning it is indeed possible to win the game at vertex $v_5$.