Suppose we have a $F[G]$-module, $M$. Note that $F[G]$ contains an isomorph of $F$ within it (as the elements $\alpha 1_G$). So for our underlying abelian group we can still use $M$, and for an element $m \in M$, and $\alpha \in F$, we can define:
$\alpha m = (\alpha 1_G)m$, where the right-hand side is the $F[G]$-action: $F[G] \times M \to M$.
It should be clear from this definition that this does indeed define an $F$-module structure on $M$ (the module axioms are readily verified, as a direct consequence of $M$ being an $F[G]$-module). Now an $F$-module is a vector space over $F$ (modules and vector spaces share the same axioms, you see, so the preceding is true by definition).
So our vector space is just $M$, as you conjectured. So how to get a representation out of this? We define $\rho(g)$ to be the mapping $m \mapsto (1_Fg)m$. This mapping is invertible, with inverse $\rho(g)^{-1}$ being the map $m \mapsto (1_Fg^{-1})m$ (we are appealing to the fact that $(1_F1_G)$ is the unit of the ring $F[G]$, so that $(1_F1_G)m = m$).
There are a couple of details left to verify: we need to show that $\rho$ is a homomorphism, and that $\rho(g)$ is $F$-linear.
Let m,m' \in M, \alpha,\beta \in F. Then \rho(g)(\alpha m + \beta m') = \rho(g)((\alpha 1_G)m + (\beta 1_G)m')
= (1_Fg)((\alpha 1_G)m + (\beta 1_G)m) = (1_Fg)((\alpha 1_G)m) + (1_Fg)((\beta 1_G)m')
= ((1_Fg)(\alpha 1_G))m + ((1_Fg)(\beta 1_G))m' = ((\alpha 1_G)(1_F g))m + ((\beta 1_G)(1_F g))m'
= (\alpha 1_G)((1_Fg)m) + (\beta 1_G)((1_Fg)m') = (\alpha 1_G)(\rho(g)(m)) + (\beta 1_G)(\rho(g)(m'))
= \alpha(\rho(g)(m)) + \beta(\rho(g)(m')), as desired, $\rho(g)$ is $F$-linear.
For arbitrary $g,h \in G, m \in M,\ \rho(gh)(m) = (1_Fgh)m = ((1_Fg)(1_Fh))m = (1_Fg)((1_Fh)m)$
$= \rho(g)((1_Fh)m) = \rho(g)(\rho(h)(m)) = (\rho(g) \circ \rho(h))(m)$, so $\rho(gh) = \rho(g) \circ \rho(h)$, thus $\rho$ is a homomorphism from $G$ to $GL_F(M)$.