Assume we have two sequences of random elements $X_{n}$ and $Y_{n}$ on some normed space $(S, \|\cdot\|_S)$ and defined on the same probability space such that $$ X_{n} \stackrel{d}{\to} \phi(Z) $$ and $$ Y_{n} \stackrel{d}{\to} \psi(Z) $$ for some r.v. $Z$ and some continuous operators $\phi$ and $\psi$. Next, let us construct the following sequence of r.v.: $$ T_{n} = X_{n} I\{X_{n} \in C\} + Y_{n} I\{X_{n} \in \bar{C}\}. $$
Does this mean then that $$ T_{n} \stackrel{d}{\to} \phi(Z) I\{\phi(Z) \in C \} + \psi(Z) I\{\phi(Z) \in \bar{C} \} $$ and $$ \|T_{n}\|_S \stackrel{d}{\to} \|\phi(Z)\|_S I\{\phi(Z) \in C \} + \|\psi(Z)\|_S I\{\phi(Z) \in \bar{C} \}? $$
$\def\dto{\xrightarrow{\mathrm{d}}}$This is not true in general.
Counter-example: Take $(S, \|·\|_S) = (\mathbb{R}, |·|)$, a random variable $Z \in U(0, 1)$, and$$X _n = Z \sim U(0, 1),\ Y_n = 1 - Z \sim U(0, 1), \quad \forall n \in \mathbb{N}_+ $$ then $|X_n| = X_n$, $|Y_n| = Y_n$, and$$ X_n \dto Z, \quad Y_n \dto Z. $$
Now take $C = \left( 0, \dfrac{1}{2} \right)$, then$$ T_n = Z I_C(Z) + (1 - Z) I_{C^c}(Z) \sim U\left( 0, \frac{1}{2} \right), $$ and $|T_n| = T_n$, but$$ Z I_C(Z) + Z I_{C^c}(Z) = Z \sim U(0, 1) \Longrightarrow T_n \not\dto Z I_C(Z) + Z I_{C^c}(Z). $$