Let $\{X_n\}, \{Y_n\}$ be sequences of real valued random variables converging in distribution to $X$ and $Y$ respectively. Let $f: \mathbb R^2 \to \mathbb R$ be a continuous function such that $\{f(X_n,Y_n)\}$ converges in distribution to some random variable $Z$.
Then is it true that $Z$ is identically distributed as $f(X,Y)$ ?
If this is not true in general, is it at least true for the function $f(x,y)=x+y$ ?
Not without independence assumptions (or convergence of joint distributions). Take $\{X_n\}$ i.i.d. with standard normal distribution, $Y_n=-X_n$ for all $n$ and $f(x,y)=x+y$. Then $\{X_n\} \to X_1$ and $\{Y_n\} \to X_1$ in distribution but $X_n+Y_n \to 0$ in distribution.
For your question to make sense, you need to mention that the vectors $(X,Y)$ and $(X_n, Y_n)$ exist (i.e. $X_n$ and $Y_n$ must live in the same probability space).
If you have the stronger assumption that $(X_n, Y_n)$ converges in distribution to the vector $(X,Y)$, then $f(X_n, Y_n)$ converges in distribution to $f(X,Y)$ by the continuous mapping theorem.
Without this stronger assumption, the claim does not hold, even for $f(x,y) = x + y$, as demonstrated in one of the answers in this question.
