Was going to be far too long to be a comment so here's an answer. The actual path deformation is the tricky bit. To see why we can deform the path we first note that we can firmly define our Lie group homomorphism $F$ in a neighbourhood $U \subset G$ of the identity using the local diffeomorphism $\exp$.
First, take a path $\gamma:[0,1] \to G$, $\gamma(0) = e$, $\gamma(1) = g$ and chop it up into little intervals $[t_i, t_{i+1}]$ if we choose these small enough we can make the jump from $\gamma(s)$ to $\gamma(t)$, which is just $\gamma(t)\gamma(s)^{-1}$ live inside $U$ for all $s,t$ in the interval. You can use this to define $F(g)$ as:
$$ F(g) = F(\gamma(t_n)\gamma(t_{n-1})^{-1})\dotsm F(\gamma(t_1)\gamma(t_0)^{-1}).$$
To make this rigorous you have to first show that our choice of partition into intervals doesn't matter (as long as they are small enough) but this is not too hard as any two partitions have a union and this can be seen to give the same answer as either partition.
Then given two paths we have to deform from one to the other in a certain way. This basically works by defining a homotopy between the two and slicing it. In other words, take $\gamma_0, \gamma_1$ two paths to $g$ and define a continuous $\gamma(\cdot,\cdot):[0,1]^2 \to G$ such that each $\gamma(s, \cdot)$ is a path to $g$ and $\gamma(0,\cdot) = \gamma_0$, $\gamma(1,\cdot) = \gamma_1$ (i.e. a homotopy from $\gamma_0, \gamma_1$).
Now we choose a small enough grid (let's say $N \times N$) on $[0,1]^2$ such that that for any two pairs $(s,t), (s',t')$ within two grid squares of each other, $\gamma(s,t), \gamma(s',t')$ differ by an element of $U$ (like with the partitions before but in 2D now). Note that this gives a sequence of paths between $\gamma_0, \gamma_1$ along the gridlines defined as $\phi_k := \gamma(k/N, \cdot)$ for $k = 0,1,\dots ,N$. We can split along the gridlines the other way as well and deform $\phi_k$ into $\phi_{k+1}$ in $N$ steps through a sequence of paths that agree with $\phi_k$ up to a certain gridline and then swap to agree with $\phi_{k+1}$ by the next gridline. The key here is that at each moment we are only deforming a path across two of our grid squares. Then we can choose a partition each time as the gridlines but skip the one we want to move across so something like $$0, \frac{1}{N}, \frac{2}{N}, \dots, \frac{l-1}{N}, \frac{l+1}{N}, \frac{l+2}{N}, \dots, 1.$$
We can choose the grid small enough to make sure this all works and then at each point we have chosen a partition that doesn't even notice the deformation we made so certainly $F(g)$ won't change.
To summarise and hopefully bring this back to visualisability: we define $F(g)$ by chopping a path into small enough pieces that we can move along by only elements in $U$. Then we can move between two paths by splitting the deformation up into a discrete sequence that at each stage we can change a path to the next one in a way that we can choose a partition that doesn't even notice the difference.
This proof is laid out in Hall's book on Lie groups. It makes the assumption that this is a matrix group but I think it works for all Lie groups with minimal adjustment. It has some pictures that may make this clearer
