For your first question: you cannot use any non-continuous $f$.
Suppose $f : \mathbb{R} \to \mathbb{R}$ is a measurable function which is not continuous. There exist measures $\mu_n, \mu$ such that $\mu_n \to \mu$ weakly but $\int f\,d\mu_n \not\to \int f\,d\mu$.
Proof. If $f$ is not continuous then there exist points $x_n, x \in \mathbb{R}$ with $x_n \to x$ but $f(x_n) \not\to f(x)$. Let $\mu_n = \delta_{x_n}, \mu = \delta_x$ be point masses at $x_n, x$ respectively. (That is, take $X_n = x_n, X = x$ to be constant random variables.)
For your second question: we want to know conditions on a set $\mathcal{C}$ of bounded continuous functions on $\mathbb{R}$ to guarantee the following statement: if $\int f\,d\mu_n \to \int f\,d\mu$ for all $f \in \mathcal{C}$, then $\mu_n \to \mu$ weakly.
A helpful sufficient condition is: the uniform closure of the linear span of $\mathcal{C}$ contains the set $C_c(\mathbb{R})$ of continuous compactly supported functions.
First note that if $\int f\,d\mu_n \to \int f\,d\mu$ for all $f \in \mathbb{C}$, then by linearity of the integral, the same holds for $f$ which are (finite) linear combinations of functions from $\mathcal{C}$. Also, if $f_m \to f$ uniformly and the above condition holds for all $f_m$, then a triangle inequality argument shows it also holds for $f$. So the condition holds on the closure of the span of $\mathcal{C}$.
Now if the condition holds on $C_c(\mathbb{R})$, then we get $\mu_n \to \mu$ weakly. This is a standard, but not quite trivial, fact. The proof goes something like this.
Find a large enough compact set $K$ that $\mu(K) \ge 1-\epsilon$. Choose $f \in C_c(\mathbb{R})$ that is 1 on $K$ and bounded by 1. From the fact $\int f\,d\mu_n \to \int f\,d\mu$, deduce that the sequence $\{\mu_n\}$ is tight. By Prohorov's theorem, $\{\mu_n\}$ is weakly precompact. Using the "double subsequence" trick, it now suffices to show that the only possible subsequential weak limit is $\mu$.
Suppose $\nu$ is a weak limit of some subsequence $\mu_{n_k}$. For every $f \in C_c(\mathbb{R})$, we have $$\int f\,d\nu = \lim_{k \to \infty} \int f \,d\mu_{n_k} = \int f\,d\mu.$$ Now for any open set $U$, we can find a sequence $f_m \in C_c(\mathbb{R})$ with $f_m \to 1_U$ pointwise and boundedly. By dominated convergence, $$\nu(U) = \lim_{m \to \infty} \int f_m\,d\nu = \lim_{m \to \infty} \int f_m\,d\mu = \mu(U).$$ Now use a monotone class argument to show $\nu(B) = \mu(B)$ for all Borel sets. Hence $\nu = \mu$.
Some examples of classes $\mathcal{C}$ satisfying this condition:
Compactly supported, piecewise linear functions
$C^\infty$ compactly supported functions
Lipschitz functions (by either of the above)
Another class $\mathcal{C}$ of functions that works, for different reasons, is $\{e^{itx} : t \in \mathbb{R}\}$. This is Lévy's continuity theorem. I don't think this $\mathcal{C}$ satisfies the previous sufficient condition.
A new question was posed in comments: suppose the probability measures $\mu_n, \mu$ are all absolutely continuous with respect to Lebesgue measure $m$, with densities $h_n, h$. Suppose moreover that they are all supported inside $[0,1]$. If $\mu_n \to \mu$ weakly, does it follow that $\int f\,d\mu_n \to \int f\,d\mu$ for all bounded measurable $f$?
Answer: No, it does not.
Let $E \subset [0,1]$ be a fat Cantor set, or your other favorite set which is closed, nowhere dense, and has positive Lebesgue measure. Let $h = \frac{1}{m(E)} 1_E$. For $1 \le k \le 2^n$, let $I_{n,k}$ denote the open interval $((k-1)2^{-n}, k2^{-n})$. Define $h_n$ by
$$f_n = \sum_{k=1}^{2^n} \frac{m(I_{n,k} \cap E)}{m(I_{n,k} \setminus E) m(E)} 1_{I_{n,k} \setminus E}.$$
Note that since $E$ is closed and nowhere dense, for any $n,k$ we have that $I_{n,k} \setminus E$ is open and nonempty; in particular it has positive measure. So this definition makes sense.
Let $d\mu_n = h_n\,dm$, $d\mu = h \,dm$ be the measures with the corresponding densities. Observe that $\mu_n(E) = 0$ for all $n$. If we take $f = 1_E$, which is bounded measurable (and even upper semicontinuous), then clearly $\int f\,d\mu_n = 0$ while $\int f\,d\mu = 1$. Now I claim that $\mu_n \to \mu$ weakly, which will complete the counterexample.
By construction, $\mu_n(I_{n,k}) = \frac{m(I_{n,k} \cap E)}{m(E)} = \mu(I_{n,k})$. Also observe that for $n \le m$ and $1 \le k \le 2^{-n}$ we have $$I_{n,k} = \left(\bigcup_{j=(k-1)2^{m-n}+1}^{k 2^{m-n}} I_{m,j}\right) \cup \text{ finitely many points} \tag{*}$$
and thus
$$\mu_m(I_{n,k}) = \sum_j \mu_m(I_{m,j}) = \sum_j \mu(I_{m,j}) = \mu(I_{n,k}) = \mu_n(I_{n,k})$$
where the sums are taken over the same $j$ as in (*).
Let $g : [0,1] \to \mathbb{R}$ be continuous and let $\epsilon > 0$. By uniform continuity, for all sufficiently large $n$ we have that the oscillation of $g$ on every interval of length $2^{-n}$ is less than $\epsilon$. So if we set
$$\tilde{g} = \sum_{k=1}^{2^n} g(k 2^{-n}) 1_{I_{n,k}}$$
then $|g - \tilde{g}| < \epsilon$ almost everywhere. Now note that $$\int \tilde{g}\,d\mu_n = \sum_{k=1}^{2^n} g(k 2^{-n}) \mu_n(I_{n,k}) = \sum_{k=1}^{2^n} g(k 2^{-n}) \mu(I_{n,k}) = \int \tilde{g}\,d\mu.$$
Hence
$$\left|\int g\,d\mu_n - \int g\,d\mu\right| = \left| \int (g-\tilde{g})\,d\mu_n + \int (\tilde{g}-g)\,d\mu \right| \le 2\epsilon.$$
So $\int g\,d\mu_n \to \int g\,d\mu$ and we have shown weak convergence.
Best Answer
You are very close.
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ an arbitrary continuous and bounded function.
Note that the function $h:\mathbb{R}^{2}\rightarrow\mathbb{R}$ prescribed by $\langle x,y\rangle\mapsto x+y$ is continuous so that the composition $f\circ h:\mathbb{R}^{2}\rightarrow\mathbb{R}$ is continuous.
Also $f\circ h$ is bounded, since $f$ is bounded.
Then $\lim_{n\rightarrow\infty}\mathbb{E}f\left(h\left(X_{n},Y_{n}\right)\right)=\mathbb{E}f\left(h\left(X,Y\right)\right)$ as a consequence of $\left(X_{n},Y_{n}\right)\stackrel{d}{\rightarrow}\left(X,Y\right)$.
This comes to the same as $\lim_{n\rightarrow\infty}\mathbb{E}f\left(X_{n}+Y_{n}\right)=\mathbb{E}f\left(X+Y\right)$.
This is true for any continuous and bounded function $f:\mathbb{R}\rightarrow\mathbb{R}$, so we are allowed to conclude that $X_{n}+Y_{n}\stackrel{d}{\rightarrow}X+Y$.