Let $\{g_1,\dots,g_n\}$, $\{f_1,\dots,f_m\}$ be Gröbner bases of polynomial ideals $G,F\subset \mathbb{R}[x_1,\dots,x_k]$, respectively, under some monomial ordering. When is $\{g_1,\dots,g_n,f_1,\dots,f_m\}$ a Gröbner basis of the ideal $G+F$?
By definition, we have $$\langle LT(g_1),\dots,LT(g_n)\rangle = \langle LT(G)\rangle$$ $$\langle LT(f_1),\dots,LT(f_m)\rangle = \langle LT(F)\rangle$$ where $LT(f)$ denotes the leading term of $f$. Thus, $$\langle LT(g_1),\dots,LT(g_n),LT(f_1),\dots, LT(f_m)\rangle =\langle LT(F)\rangle+\langle LT(G)\rangle.$$ I'm interested in conditions on when $\langle LT(F)\rangle+\langle LT(G)\rangle=\langle LT(F+G)\rangle$.
Consider $R=k[x]$. There is only one monomial order in this case which is same as degree. Now consider the two principal ideals $F=\langle x^2+x\rangle$ and $G=\langle x^2\rangle$. Clearly $\{x^2+x\}$ and $\{x^2\}$ are Gröbner bases for $F$ and $G$. However $$F+G=\langle x^2+x,x^2\rangle=\langle \gcd(x^2+x,x^2)\rangle=\langle x\rangle, $$ and any Gröbner basis for $\langle x\rangle$ should have $cx$ for a non-zero scalar $c\in k$ among its elements, otherwise leading terms of its elements won't generate the initial ideal of $\langle x\rangle$. Therefore clearly the set $\{x^2+x,x^2\}$ is not a Gröbner basis for $F+G$.
So we showed an example where for no monomial order the union of Gröbner bases of two ideals can be a Gröbner basis for the sum of the two ideals.