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Under $H_0$, the sequence $(T_n(\lfloor n\,\cdot\rfloor)/\sqrt n)_{n=1}^\infty$ (you forgot to divide by $\sqrt n$) converges to the Brownian bridge $B^\circ(\cdot)$ in distribution. Suppose that (as a specification of $H_1$) $h_n/\sqrt n\to h\ne0$ and $k^*/n\to\kappa\in(0,1)$, where $h_n:=EX_n-EX_1$. Since the distribution of the sequence $(X_1,\dots,X_n)$ under $H_1$ is the same as that of the sequence $$(X_1,\dots,X_{k*},X_{k^*+1}+h_n\dots,X_n+h_n)$$ $$=(X_1,\dots,X_n)+h_n\,(\underbrace{0,\dots,0}_{k^*},1,\dots,1)$$ under $H_0$, it follows that, under the mentioned specification of $H_1$, the sequence $(T_n(\lfloor n\,\cdot\rfloor)/\sqrt n)_{n=1}^\infty$ converges to $B^\circ(\cdot)+hI\{\cdot>\kappa\}$ in distribution (say in the Skorokhod space $D[0,1]$), where $I$ is the indicator function.