Counting measure is just summation!
To see this, you can approach from a few different angles; how about we consider the Monotone Convergence Theorem. To that end, for $n\in\mathbb{N}$, define $f_n:\mathbb{N}\to\mathbb{R}$ by
$$
f_n(k)=\begin{cases}f(k) & \text{if }1\leq k\leq n\\ 0 & \text{else}\end{cases}.
$$
Then clearly, as $n\to\infty$, $f_n\to f$ pointwise; it is also monotone increasing, because $f(k)\geq0$ for all $k\in\mathbb{N}$. So, by the MCT,
$$
\int f_n\,d\mu\to\int f\,d\mu\text{ as }n\to\infty.
$$
Now, consider these $f_n$. Note that we can write
$$
\mathbb{N}=\{1\}\cup\{2\}\cup\cdots\cup\{n\}\cup\{n+1,n+2,\ldots\},
$$
and that these sets are all measurable. So,
$$
\begin{align*}
\int f_n\,d\mu&=\int_{\{1\}}f_n\,d\mu+\cdots+\int_{\{n\}}f_n\,d\mu+\int_{\{n+1,n+2,\ldots\}}f_n\,d\mu\\
&=\int_{\{1\}}f_n(1)\,d\mu+\cdots+\int_{\{n\}}f_n(n)\,d\mu+\int_{\{n+1,n+2,\ldots\}}0\,d\mu,
\end{align*}
$$
where we have used that $f_n$ is constant on each of these sets, by definition. So, we see that
$$
\int f_n\,d\mu=1\cdot f_n(1)+1\cdot f_n(2)+\cdots+1\cdot f_n(n)+0=f(1)+\cdots+f(n).
$$
So, we have that
$$
\int f\,d\mu=\lim_{n\to\infty}\int f_n\,d\mu=\lim_{n\to\infty}(f(1)+\cdots+f(n))=\sum_{k=1}^{\infty}f(k).
$$
The boundedness is of no consequence here -- since our terms are non-negative, the series either converges or diverges to $\infty$; in either case, the integral is exactly the sum.
Best Answer
First note that by definition,
$$\int_{\{x\}} f\;d\mu=\int f\cdot1_{\{x\}}d\mu.$$
However, note that $f\cdot 1_{\{x\}}=f(x)1_{\{x\}}$. Therefore we use linearity of the integral to compute
$$\int_{\{x\}}f\;d\mu=\int f(x)1_{\{x\}}d\mu=f(x)\int 1_{\{x\}}d\mu=f(x)\,\mu(\{x\})=f(x)$$
where the last equality is by definition of the counting measure.
I'd recommend working on the more general case on your own, using what we just proved.