There is a little step on the proof of Van Kampen's theorem that I am missing. I will adapt the argument so that I can expose my problem as an isolated exercise:
Let $\omega$ and $\eta$ be two paths along a topological space $X$ which join a point $x$ to a point $y$. Suppose $X=U\cup V$ where $U$, $V$ and $U\cap V$ are path-connected open subsets. Suppose that $\omega$ and $\eta$ are path homotopic and let $F:\left [0,1\right ]\times \left [0,1\right ]\to X$ such that $F\left (s,0\right )=\omega \left (s\right )$, $F\left (s,1\right )=\eta \left (s\right )$, $F\left (0,t\right )=x$ and $F\left (1,t\right )=y$.
Suppose also that there exists a partition $0=t_0<\cdots <t_n=1$ of $I=\left [0,1\right ]$ such that if $I_i=\left [t_{i-1},t_i\right ]$ then $F\left (I_i\times I\right )$ is contained either in $U$ or $V$.
For every $i$ let $u_i:I\to \left [t_{i-1},t_i\right ]$ such that $u_i\left (s\right )=\left (1-s\right )t_{i-1}+st_i$ and call $\omega_i:=\omega \circ u_i$, $\eta_i:=\eta\circ u_i$. Also, define $\beta_i:I\to X$ such that $\beta_i\left (t\right )=F\left (t_i,t\right )$. We have to prove that $\left [\beta_{i-1}\ast \eta_i\right ]=\left [\omega_i\ast \beta_i\right ]$ as paths either in $U$ or $V$.
I really do not know how to prove it, nor even how to begin. I tried defining $H\left (s,t\right )=F\left (u_i\left (s\right ),t\right )$. Observe that $H\left (s,0\right )=\omega_i\left (s\right )$, therefore a path homotopy between $\beta_{i-1}\ast \eta_i$ and $\omega_i\ast \beta_i$ should (maybe) be defined as $G\left (s,t\right )=H\left (2s,t\right )$ for $s\leq \frac{1}{2}$. But I do not know how to completely define $G$, nor whether this is the correct way to define the first part of the homotopy. How would you prove the assertion?
At the moment your subdivision arguments involve only paths, whereas to deal with homotopies you need to subdivide a square into little squares, each of which lies in either $U,V$, or both, and then join images of the vertices of this subdivision in $X$ to base points staying in either $U,V$ or $U \cap V$, by the connectivity assumption. Then the whole square has to be deformed to a subdivided square in which each small square is mapped to $U$ or $V$ and all vertices to base points.
More details are in this paper, and background also in this one.