Definition of addition Given $\alpha,\beta$ ordinal numbers, then $$ \alpha+\beta=\operatorname{ord}\big(\{0\}\times\alpha\cup\{1\}\times\beta\big). $$
Lemma Given three ordinals $\alpha$, $\beta$, and $\gamma$, then $$ \alpha<\beta \to \gamma+\alpha<\gamma+\beta. $$
Proof Given $\alpha<\beta$, then $$\tag{1} \alpha\subsetneq\beta, $$ $\alpha$ is an proper initial segment of $\beta$, thus \begin{gather*}\tag{2} \{0\}\times \gamma\cup \{1\}\times\alpha \subsetneq \{0\}\times \gamma\cup \{1\}\times\beta, \end{gather*} and the l.h.s. is an initial segment of the r.h.s., thus the claim follows: $$\tag{3} \gamma+\alpha =\operatorname{ord}\left(\{0\}\times \gamma\cup \{1\}\times\alpha\right) < \operatorname{ord}\left(\{0\}\times \gamma\cup \{1\}\times\beta\right) =\gamma+\beta. $$
Let $\langle A,<_R\rangle,\;\langle B,<_S\rangle$ and $\langle C,<_T\rangle$ be three well ordered sets with respective ordinals $\alpha,\beta$ amd $\gamma$ such that the sets $A,B$ and $C$ are pairwise disjoint. Since $\alpha<\beta$, then the well-ordered set $\langle A,<_R\rangle$ is isomorphic to an initial section $\text{sec}(b,B,<_S)$ of $\langle B,<_S\rangle$.
In the ordered sum $\langle C\cup B,<_{T\oplus S}\rangle$, the initial section $\text{sec}(b,C\cup B,<_{T\oplus S})$ determined by the element $b$ of $B$ is equal to the ordered sum $\langle C\cup A,<_{T\oplus R}\rangle$, so the ordinal $\gamma+\alpha$ is strictly less than the ordinal $\gamma+\beta$ of $\langle C\cup B,<_{T\oplus S}\rangle$