Let $X \in \mathcal{S}_+^3$ be a $3 \times 3$ positive semidefinite matrix, and let $\mathcal{A}$ be a linear transformation from $\mathcal{S}^3 \rightarrow \mathbb{R}^3$ defined as follows: \begin{equation} \mathcal{A}(X) = \left(\begin{array}{c} X_{11} \\ X_{33} \\ X_{22} + X_{13} \end{array} \right) \end{equation}
Why is the image of $\mathcal{S}_+^3$ under $A$ a union of $\mathbb{R}_+^3$ with a rotated copy of $\mathcal{S}_+^2$ ?, i.e., why is the following true? \begin{equation} \mathcal{A}(\mathcal{S}_+^3) = \mathbb{R}³_+ \cup \{(x,y,z)|x\geq 0, y \geq 0, z \leq 0, xy \geq z^2 \} \end{equation}
Let $S$ denote the set $S = \mathbb{R}³_+ \cup \{(x,y,z)|x\geq 0, y \geq 0, z \leq 0, xy \geq z^2 \}$.
By considering the matrices $$ X = \pmatrix{x & 0 & 0\\0 & z & 0\\0 & 0 & y}, \quad \pmatrix{x & 0 & z\\0 & 0 & 0\\z & 0 & y}, $$ we can see that $S \subseteq \mathcal A(\mathcal S_+^3)$.
For the reverse containment, it suffices to show that for $X \in \mathcal S_+^3$: if $\mathcal A(X) \notin \Bbb R^3_+$, then $\mathcal A(X) \in \{(x,y,z)|x\geq 0, y \geq 0, z \leq 0, xy \geq z^2 \}$.
So, suppose that $\mathcal A(X) \notin \Bbb R^3_+$. For any $X \in \mathcal S_+^3$, we can see that $X_{11},X_{22},X_{33} \geq 0$. Thus, the fact that $\mathcal A(X) \notin \Bbb R^3_+$ implies that $X_{13} + X_{22} < 0$. Because $X_{22} \geq 0$, this can only occur if $X_{13} < 0 \leq X_{22}$, which means that $|X_{13} + X_{22}| \leq |X_{13}|$. Because $X$ is positive semidefinite, its principal submatrix $$ \pmatrix{X_{11}&X_{13}\\X_{13} & X_{33}} $$ must be positive semidefinite, so that $X_{11}X_{33} \geq X_{13}^2 \geq (X_{13} + X_{22})^2$. So, it indeed holds that $\mathcal A(X) \in \{(x,y,z)|x\geq 0, y \geq 0, z \leq 0, xy \geq z^2 \}$.