$G(x_1,\dots, x_n)\in K[x_1,\dots, x_n]$ is a homogeneous of degree $d$ polynomial over algebraically closed field $K$. $[c_i,d_i]\in P^1_K$ where $P^1_K$ is indicating $c_i,d_i$ cannot be simultaneously $0$ for each $i$, then $G(c_1u+d_1v,\dots, c_nu+d_nv)\in K[u,v]$ is degree $d$?
This cannot be true for any $\forall i,[c_i,d_i]\in P^1_K$ and all degree $d$. It is totally possible to have $G(c_1u+d_1v,\dots, c_nu+d_nv)=0\in K[u,v]$ for some good choice of $[c_i,d_i]\in P^1_K$. The counting of degree freedom shows that asymptotically for large degree $d$, I cannot cancel out all coefficient by appropriate choice of $[c_i,d_i]$. However for low degree and large $n$, I could make $G(c_1u+d_1v,\dots, c_nu+d_nv)=0\in K[u,v]$.
$\textbf{Q:}$ Is above correct? If I am wrong, please provide hint to see why $G(c_1u+d_1v,\dots, c_nu+d_nv)\in K[u,v]$ necessarily non-vanishing.
Not true, just like you wrote. For simplicity we write $x,y,z$ instead of $x_1,x_2,x_3$ and say $K=\mathbb C$. Suppose then that $G(x,y,z)=x^2+y^2-2z^2$ and choose : \begin{align*} (c_1:d_1) = (1:0)\\ (c_2:d_2) = (1:0)\\ (c_3:d_3) = (1:0)\\ \end{align*}
then $G(u,u,u) = 0$