$Ric$ is Ricci curvature. Making orthogonal decomposition, there is $$ Ric =\mathring{Ric} + \frac{R}{n}g \tag{1} $$ where $R$ is scalar curvature, $g$ is Riemannian metric. I get (1) from the 42th page of Topping's Lectures on the Ricci flow.
First, I don't know why the coefficient of $g$ is $\frac{R}{n}$.
Besides, Topping state that $$ |Ric|^2 = |\mathring{Ric}|^2 + \frac{R^2}{n^2} |g|^2 \ge 0+\frac{R^2}{n} $$ Why $\frac{R^2}{n^2} |g|^2 \ge \frac{R^2}{n}$ ?
The traceless Ricci curvature $\mathring{\mathrm{Ric}}$ is defined as $$\mathring{\mathrm{Ric}} = \mathrm{Ric} - \frac1nRg .$$ The reason that we the coefficient is $\frac1nR$ is because this is precisely the coefficient that makes the traceless Ricci curvature have zero trace: $$\mathrm{tr}_g\, \mathring{\mathrm{Ric}} =\mathrm{tr}_g\, \mathrm{Ric} - \frac1n R \,\mathrm{tr}_g\,g=R-R=0.$$
As for the inequality, you have $$ \vert\mathrm{Ric} \vert^2 = \vert \mathring{\mathrm{Ric}} \vert^2 + \frac{R^2}{n^2} \vert g \vert^2 $$ since the decomposition (1) is orthogonal. As Gunnar mentions in the comments, we can calulate $\vert g \vert^2 $ directly. In coordinates the norm squared of a $(0,2)$-tensor $T_{ij}$ is $$ \vert T \vert^2 = g^{ik}g^{j\ell}T_{ij}T_{k\ell}. $$ Here $T_{ij}=g_{ij}$ so \begin{align*} \vert g \vert^2 = g^{ik}g^{j\ell}g_{ij}g_{k\ell} =\delta^k_j \delta^j_k=\delta^k_k=n. \end{align*} Hence, $$ \frac{R^2}{n^2} \vert g\vert^2=\frac{R^2}{n}. $$