The $n$-dimensional cross-polytope:
$$C_n=\{x\in \mathbf{R}^n : \pm x_1 \pm x_2 \cdots \pm x_n \leq 1\}$$
How to realize $C_n$ is the projection on the $x$-coordinate of the polytope: $$Q_n=\{(x,y)\in \mathbf{R}^{2n} :\sum_{i=1}^n y_i=1,-y_i\leq x_i \leq y_i, \forall i = 1,\ldots,n\}$$
How to understand this from the definition of projection?
It looks like we can sum them up:
$$-(\sum y_i)\leq \sum x_i \leq \sum y_i \Rightarrow -1 \leq \sum x_i \leq 1$$
However, I still cannot understand why $C_n$ is the projection of $Q_n$
The linear map $\pi : \mathbb{R}^{2n} \to \mathbb{R}^{2n}$, $(x,y) \mapsto (x,0)$ is referred to as a $\textit{projection}$, a quick google search gives you the general definition, and it is immediate to check that $\pi$ indeed satisfies $\pi^2=\pi$. When we say that $C_n$ is the projection of $Q_n$ under $\pi$, we mean that $\pi(Q_n)=C_n$. As pointed out in the comments, you can prove this equality by double inclusion.
As to the origin of the term, imagine in the case $n=1$, that a light illuminates the plane from above, and consider a point $p:=(x,y)$. Then, the point $(x,0)=\pi(p)$ is the "shadow" of $p$ on the line $y=0$ under the light, that is, its "projection". Does this help?