One such condition is that $f$ be absolutely continuous in $[0,h)^2$ for some $h\in(0,1)$ -- so that $$f(x,y)+f(0,0)-f(x,0)-f(0,y)=\int_0^x du\,\int_0^y dv\,g(u,v)$$ for some function $g$ integrable on $[0,h)^2$ and for all $(x,y)\in[0,h)^2$ -- with $g$ continuous on the set $([0,h)\times\{0\})\cup(\{0\}\times[0,h))\subset[0,h)^2$. (In particular, it suffices that $\partial_x\partial_y f$ exist on $(0,h)^2$ and admit a continuous extension to $[0,h)^2$.)

Indeed, then, by dominated convergence, for all $(x,y)\in[0,h)^2$ $$(\partial^+_y f)(x,0)=(\partial^+_y f)(0,0)+\int_0^x du\,g(u,0)$$ and hence $$(\partial^+_x \partial^+_y f)(0,0)=g(0,0).$$ Similarly, $(\partial^+_y \partial^+_x f)(0,0)=g(0,0)$. So, $(\partial^+_x \partial^+_y f)(0,0)=(\partial^+_y \partial^+_x f)(0,0)$, as desired.

One such condition is that $f$ be absolutely continuous in $[0,h)^2$ for some $h\in(0,1)$ -- so that $$f(x,y)+f(0,0)-f(x,0)-f(0,y)=\int_0^x du\,\int_0^y dv\,g(u,v)$$ for some function $g$ integrable on $[0,h)^2$ and for all $(x,y)\in[0,h)^2$ -- with $g$ continuous on the set $([0,h)\times\{0\})\cup(\{0\}\times[0,h))\subset[0,h)^2$. (In particular, it suffices that $\partial_x\partial_y f$ exist on $(0,h)^2$ and admit a continuous extension to $[0,h)^2$.)

Indeed, then, by dominated convergence, for all $x\in[0,h)$ $$(\partial^+_y f)(x,0)=(\partial^+_y f)(0,0)+\int_0^x du\,g(u,0)$$ and hence $$(\partial^+_x \partial^+_y f)(0,0)=g(0,0).$$ Similarly, $(\partial^+_y \partial^+_x f)(0,0)=g(0,0)$. So, $(\partial^+_x \partial^+_y f)(0,0)=(\partial^+_y \partial^+_x f)(0,0)$, as desired.