Given two paths $f,g:\mathbb{I}\rightarrow X$ with $f\left(1\right)=g\left(0\right)$ there is a composite $f.g$ defined by $t\mapsto f\left(2t\right)$ if $2t\leq1$ and $t\mapsto g\left(2t-1\right)$ otherwise.
Would in not be better if this would define $g.f$ instead of $f.g$?
Going out from $x=f\left(0\right)$, $f\left(1\right)=y=g\left(0\right)$ and $z=g\left(1\right)$, we have $f$ and $g$ as representatives of arrows $\left[f\right]:x\rightarrow y$ and $\left[g\right]:y\rightarrow z$ in fundamental groupoid $\pi\left(X\right)$. This gives the composite $\left[g\right]\left[f\right]:x\rightarrow z$.
In my view covariant functors (here $\left[\right]:\mathbf{Top}\rightarrow\mathbf{Grpd}$) are more natural than contravariant functors.
If you prefer to choose composition from right to left in abstract categories, then indeed, it would be much more convenient to define the composition of paths from right to left again, i.e. $g.f$ in your question.