Suppose $v=\alpha + \beta u$ for $\beta > 0$ i.e. $v$ is a positive affine transformation of $u$, then
$$ \frac{v(g^1) - v(g^2)}{v(g^2) - v(g^3)} = \frac{\alpha + \beta u(g^1) - \alpha - \beta u(g^2)}{\alpha + \beta u(g^2) - \alpha - \beta u(g^3)} = \frac{u(g^1) - u(g^2)}{u(g^2) - u(g^3)}$$
Converse is not true because $v=-u$ also satisfy
$$ \frac{v(g^1) - v(g^2)}{v(g^2) - v(g^3)} = \frac{u(g^1) - u(g^2)}{u(g^2) - u(g^3)}$$
for all $g_1, g_2, g_3\in\mathcal{G}$.
Added Later
By the way, converse is also true. We are given this information that $u$ and $v$ both represent $\succsim$. So, $v=-u$ is not a valid example as they cannot both represent $\succsim$.
Here is the proof of the converse:
Fix any two lotteries $b$ and $w$ such that $u(b) > u(w)$ and consequently, $v(b) > v(w)$. For any lottery $g\in\mathcal{G}$, we know that the following is true
$$ \frac{v(g) - v(w)}{v(w) - v(b)} = \frac{u(g) - u(w)}{u(w) - u(b)}$$
So, $$v(g) = v(w) + (v(w) - v(b))\frac{u(g) - u(w)}{u(w) - u(b)} $$
or equivalently,
\begin{eqnarray*} v(g) & = & v(w) - \frac{v(w) - v(b)}{u(w) - u(b)}u(w)+ \frac{v(w) - v(b)}{u(w) - u(b)}u(g) \\ & = & \left(\frac{u(w)v(b) - v(w)u(b)}{u(w) - u(b)}\right)+ \left(\frac{v(w) - v(b)}{u(w) - u(b)}\right)u(g) \end{eqnarray*}
Therefore, $v = \alpha + \beta u$ where $\displaystyle\alpha = \frac{u(w)v(b) - v(w)u(b)}{u(w) - u(b)}$ and $\displaystyle\beta = \frac{v(w) - v(b)}{u(w) - u(b)} > 0$.