Let $A$ be a ring, $M$ a graded $A$-module, $N$ a graded submodule of $M$ and $(M_\lambda)_{\lambda\in\Delta}$ the grading on $M$. Then $((M_\lambda+N)/N)_{\lambda\in\Delta}$ is a grading of $M/N$.
Why do we need $N$ to be graded? Let $\eta:M\rightarrow M/N$ be the canonical $A$-linear surjection.
If $i_\lambda:M_\lambda+N\rightarrow M$ is the canonical $\mathbb{Z}$-linear injection and $\pi_\lambda:M_\lambda+N\rightarrow(M_\lambda+N)/N$ the canonical $\mathbb{Z}$-linear surjection, then there exists a unique $\mathbf{Z}$-linear injection $\bar{i}_\lambda:(M_\lambda+N)/N\rightarrow M/N$ such that $\eta=\bar{i}_\lambda\circ\pi_\lambda$. Furthermore there exist a unique $\mathbb{Z}$-linear injection $$j:\bigoplus_{\lambda\in\Delta}\text{Im}(\bar{i}_\lambda)\rightarrow M/N.$$ Clearly this map is surjective. Moreover, we have $A_\lambda\text{Im}(\bar{i}_\mu)\subset\text{Im}(\bar{i}_{\lambda+\mu})$. So, where is the assumption that $N$ is graded used?
Edit: Obviously, the quoted text is using $\text{Im}(\bar{i}_\lambda)\cong (M_\lambda+N)/N$.
Your map $j$ is surjective, but how do you know it is injective? You need to know that if you take elements $a_\lambda\in M_\lambda$ and their sum $\sum a_\lambda$ is an element of $N$ (so $j$ maps $(\pi_\lambda(a_\lambda))$ to $0$), then actually each $a_\lambda$ was in $N$. This means exactly that $N$ is graded, since it is saying that if $a=\sum a_\lambda\in N$ then each of the homogeneous parts $a_\lambda$ of $a$ are in $N$ as well.