Assume $f \in K[x_1,\dots ,x_m],x_i\in F$
$$p \in K[x_1,\dots, x_m], p = \sum_{i}\phi_ix_1^{{a}_{i1}}x_2^{a_{i2}}\dots x_n^{a_{in}}, L(p) := \phi _lx^l : x^a \prec x^l \forall a\in A(p) \backslash\{l\}$$
$$\textbf{th. 1}: [\forall f_i, f_j \in\{f_1,f_2 \dots,f_n\} = F : \nexists \lambda \in F: L(f_i) = \lambda L(f_j) ] \rightarrow [F - \text{Grobner basis}] (\textbf{It isn't true, following Marcus Ritt's counterexample})$$
$$\textbf{th. 2}: [\forall f_i, f_j \in\{f_1,f_2 \dots,f_n\} = F : \forall i \neq j : L(f_i) \text{ and } L(f_j) \text{ are mutually prime} ] \rightarrow [F - \text{Grobner basis}]$$
Looks like this feature works in general, but i don't know, how to prove it. Is it true, that if left-hand condition is true, then $S$-polynomial $S(f_i,f_j) \overset{F}{\rightsquigarrow} 0$ (reducible to zero), so i could use Buchberger criteria ?
No, for example $F=\{f_1,f_2\}$ with $f_1=x^2y-2y^2+x$ and $f_2=x^3-2xy$ satisfies your criterion, but is not a Gröbner basis, since $S(f_1,f_2)=x^2$.