Your original understanding of the lexicographic order on $\kappa^{<\omega}$ is correct. The claim isn’t true as stated, but something that should be good enough for the intended purpose is true.
Proposition. Each ordinal $\alpha<\kappa^+$ can be order-embedded in $\kappa^{<\omega}$.
Proof. Suppose that $\varphi:\alpha\to\kappa^{<\omega}$ is an order embedding. Then
$$\varphi^+:\alpha+1\to\kappa^{<\omega}:\xi\mapsto\begin{cases}
0^\frown\varphi(\xi),&\text{if }\xi\in\alpha\\
\langle 1\rangle,&\text{if }\xi=\alpha
\end{cases}$$
is an order-embedding of $\alpha+1$. If $\alpha<\kappa^+$ is a limit ordinal, and each $\beta<\alpha$ can be order-embedded in $\kappa^{<\omega}$, let $\lambda=\operatorname{cf}\alpha\le\kappa$, and let $\langle\alpha_\xi:\xi<\lambda\rangle$ be a strictly increasing sequence cofinal in $\alpha$ such that $\alpha_0=0$. For $\xi<\lambda$ let $\beta_\xi$ be the order type of the interval $I_\xi=[\alpha_\xi,\alpha_{\xi+1})$; clearly $\beta_\xi\le\alpha_{\xi+1}<\alpha$, so there is an order-embedding $\varphi_\xi$ of $I_\xi$ into $\kappa^{<\omega}$. For $\eta\in\alpha$ let $\xi(\eta)<\lambda$ be such that $\eta\in I_{\xi(\eta)}$, and let
$$\varphi:\alpha\to\kappa^{<\omega}:\eta\mapsto\xi(\eta)^\frown\varphi_{\xi(\eta)}(\eta)\;;$$
then $\varphi$ order-embeds $\alpha$ in $\kappa^{<\omega}$. $\;\dashv$
Now let $x,y\in\kappa^{<\omega}$ be such that $x\prec y$, where $\preceq$ is the lexicographic order on $\kappa^{<\omega}$. Assume further that if $y=x^\frown z$, at least one term of $z$ is not $0$. Then there is an $x'\succeq x$ such that $x'\prec y$ and $x'$ is not an initial segment of $y$. Let
$$I=\left\{z\in\kappa^{<\omega}:x'\subsetneqq z\right\}\;$$
then $I$ is order-isomorphic to $\kappa^{<\omega}$, and $I\subseteq(x',y)\subseteq(x,y)$. Thus, each ordinal less than $\kappa^+$ order-embeds into any interval $(x,y)$ of $\langle\kappa^{<\omega},\preceq\rangle$ such that $y$ is not an extension of $x$ by a string of zeroes.
