Given the curve $\gamma=(\gamma_1(t),\gamma_2(t))$, with
$$\gamma_1(t) = \cases {t & $0<t\leq1$\\ 3-t & $2<t\leq3$\\ 0 & otherwise}$$ and $$\gamma_2(t) = \cases {t-1 & $1<t\leq2$\\ 4-t & $3<t\leq4$\\ 0 & otherwise}$$ this curve surrounds a unit area square, with the bottom-left corner in the origin. I want to verify that the antisymmetric component of the path signature element $S^2_{0,4}(\gamma)$ is equal to area enclosed by the square, that means:
$$\text{Anti}\left[S^2_{0,4}(\gamma)\right]=\frac12\left(\int_{\Delta_{[0,4]}}\text d\gamma_1\text d\gamma_2-\int_{\Delta_{[0,4]}}\text d\gamma_2\text d\gamma_1\right)=1,$$ where $\Delta_{[0,4]}$ designates the 2-dimensional symplex.
First of all, I proceed to compute:
$$\int_{\Delta_{[0,4]}}\text d\gamma_1\text d\gamma_2=\int_0^1\int_0^{t_2}\dot\gamma_1\text dt_1\,\dot \gamma_2\text dt_2+\int_1^2\int_1^{t_2}\dot\gamma_1\text dt_1\,\dot \gamma_2\text dt_2+\dots,$$ where the dotted quantities represent the components of the velocity.
Since on every side of the square one of the components of the velocity is zero, I obtain that all the integrals I would write would be null, and similarly the enclosed area. What am I missing?
I presume you are missing that $\gamma_1(t)=1$ for $1<t\le2$ and $\gamma_2(t)=1$ for $2<t\le3$, otherwise you don't have a square or even a continuous path.
The iterated integrals which define a path signature are Riemann-Stieltjes integrals. When the path is not differentiable, you need to interpret them as such, and not try to expand them as derivatives (which do not exist). Instead of thinking of $\int_0^{t_2}d\gamma_1(t_1)$ as $\int_0^{t_2}\dot\gamma_1\,dt_1$, you can just evaluate it directly as $\gamma_1(t_2)-\gamma_1(0)=\gamma_1(t_2)$.
We can then try to evaluate the outer integral $$\int_0^4\int_0^{t_2}d\gamma_1(t_1)d\gamma_2(t_2)=\int_0^4 [\gamma_1(t_2)] \,d\gamma_2(t_2)$$ Such integrals are zero on intervals where $\gamma_2$ is constant, so this is $$\int_1^2 \gamma_1(t_2) \,d\gamma_2(t_2)+\int_3^4 \gamma_1(t_2) \,d\gamma_2(t_2)=\int_1^2 1 \,d\gamma_2(t_2)+\int_3^4 0 \,d\gamma_2(t_2)$$ which is $$\gamma_2(2)-\gamma_2(1)=1-0=1$$