If $(X, x_0)$ is an H-space, then $\pi_1(X,x_0)$ is Abelian.
I started with knowing I had to shows $[f][g]=[g][f]$, which means showing $f * g \simeq g * f \text{ rel } \{0,1\}$ but I contructed a homotopy from this by partitioning the unit square into $6$ triangles made from drawing in both diagonals and a vertical line from $(\frac{1}{2}, 1)$ to $(\frac{1}{2}, 0)$, where: LL means lower left triangle, UR means upper right triangle, L means left triangle, etc...
But I realized that I didn't use the fact that the space is an H-space at all.
$F(s,t) = \begin{cases} f(2\frac{t-s}{2t-1}), & \text{if $(s,t)$ in LL} \\ g(\frac{2s-1}{1-2t}), & \text{if $(s,t)$ in LR} \\ x_0, & \text{if $(s,t)$ in L or R} \\ g(2\frac{1-s-t}{1-2t}), & \text{if $(s,t)$ in UL} \\ f(\frac{1-2s}{1-2t}), & \text{if $(s,t)$ in UR} \\ \end{cases}$
So given this homotopy is continuous (all of its inner functions are continuous since they're all affine maps in their respective domains), what did I assume incorrectly?
Here's the diagram I used:
Your $F$ is not continuous at $(s,t)=(\frac12,\frac12)$ -- for example, there are points arbitrarily close to $(\frac12,\frac12)$ where $F$ equals $f(\frac12)$, which is generally some nonzero distance from $x_0$.
You're probably confusing yourself by the fact that the restriction of $F$ to any horizontal or vertical line is continuous -- but that doesn't mean that $F$ itself is continuous.