While I'm sure that the first part of the proof is correct, I'm unable to verify the second one. Any help to verify that part is greatly appreciated.
My attempt:
Let $(W_1,<_1)$ and $(W_2,<_2)$ be well-ordered sets and isomorphic to $\alpha_1$ and $\alpha_2$ respectively, and let $(W,<)$ be the sum of $(W_1,<_1)$ and $(W_2,<_2)$. Then $(W,<)$ isomorphic to the ordinal $\alpha_1+\alpha_2$.
We assume that $W_1\cap W_2=\emptyset$ and $W_1\cup W_2=W$, and that each element in $W_1$ precedes in $<$ each element of $W_2$, while $<$ agrees with $<_1$ and with $<_2$ on both $W_1$ and $W_2$. We prove the theorem by induction on $\alpha_2$.
- If $\alpha_2=0$, then $W_2=0$, $W=W_1$, and $\alpha_1+\alpha_2=\alpha_1$.
The below part is too complicated for me to verify. Please help me check it out!
If $\alpha_2=\beta+1$, then $W_2$ has the greatest element $a$. By IH, $W_1\cup W_2[a] \cong \alpha_1+\beta$ where $W_2[a]:=\{x\in W_2\mid x<_2 a\}$. We extend the isomorphism between $W_1\cup W_2[a]$ and $\alpha_1+\beta$ to an isomorphism between $W_1\cup W_2$ and $(\alpha_1+\beta)+1$ by mapping $a$ to $\alpha_1+\beta$. Moreover, $(\alpha_1+\beta)+1=\alpha_1+(\beta+1)=\alpha_1+\alpha_2$. Thus $W \cong \alpha_1+\alpha_2$.
If $\alpha_2$ is a limit ordinal. For each $\beta<\alpha_2$, there is a unique isomorphism $f_{\beta}$ between $W_1\cup W_2[a_{\beta}]$ and $\alpha_1+\beta$ where $a_{\beta}$ is the $\beta^{\text{th}}$ element of $W_2$ and $W_2[a_{\beta}] \cong\beta$. It's clear that $\beta_1<\beta_2\implies f_{\beta_1} \subsetneq f_{\beta_2}$. Let $f=\bigcup_{\beta<\alpha_2}f_{\beta}$. Then $f$ is an isomorphism between $W_1\cup\left(\bigcup_{\beta<\alpha_2} W_2[a_{\beta}]\right)=W_1\cup W_2=W$ and $\bigcup_{\beta<\alpha_2}(\alpha_1+\beta)$. Moreover, $\alpha_1+\alpha_2=\sup\{\alpha_1+\beta\mid \beta<\alpha_2\}=\bigcup_{\beta<\alpha_2}(\alpha_1+\beta)$. It follows that $f$ is an isomorphism between $W$ and $\alpha_1+\alpha_2$, or equivalently $W\cong \alpha_1+\alpha_2$.
Edit: I added the definition of the sum of orders here.
Let $(W_1,<_1)$ and $(W_2,<_2)$ be linearly ordered sets and $W_1 \cap W_2=\emptyset$. The relation $<$ on $W=W_1 \cup W_2$ defined by $$a<b \text{ if and only if } a,b\in W_1 \text{ and } a<_1 b $$ $$\text{or } a,b\in W_2 \text{ and } a<_2 b $$ $$\text{or } a\in W_1,b\in W_2$$ Then $(W,<)$ is called the sum of $(W_1,<_1)$ and $(W_2,<_2)$.
In the limit step you need to prove that, if $f_k$ is an order isomorphic between two wosets $X_k$ and $Y_k$, where $k$ is an element of a woset $I$, and: $$ \forall m,n \in I: [m \preceq_I n \Rightarrow (X_m \subseteq X_n \wedge Y_m \subseteq Y_n)] $$ then $\bigcup f_k$ is an order isomorphic between $\bigcup X_k$ and $\bigcup Y_k$.
After that, you can apply for order isomophic $f_k$ between $\bigcup [(W_1, \preceq_1) \oplus (s_{W_2}(k), \preceq_2)]$ and $\bigcup (\alpha_1 + \beta)_{\beta \in \alpha_2}$.