Being measurable w.r.t. $m\times\nu$ doesn't make sense, and furthermore you don't even use the product measure in this exercise.
Instead, you should have specified which sigma-algebra you equip $[0,1]$ with - both when you're speaking of $m$ and when you're speaking of $\nu$. You could for example consider the measure-space $([0,1],\mathcal{E},m)$ and $([0,1],\mathcal{F},\nu)$, where $\mathcal{E}=\mathcal{B}([0,1])$ is the Borel sigma-algebra on $[0,1]$ and $\mathcal{F}$ could be $\mathcal{B}([0,1])$ or even the power set $\mathcal{P}([0,1])$. But let us assume that $\mathcal{F}=\mathcal{B}(\mathbb{R})$ since this is the smallest of the two.
Now you should show that $$D\in\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})=\mathcal{B}(\mathbb{R}^2),$$
i.e. that $D$ belongs to the product-sigma-algebra of $[0,1]\times [0,1]$. One strategy for that is to show that $D$ is closed in $\mathbb{R}^2$. This ensures that the sections $D_x=\{y\in\mathbb{R}\mid (x,y)\in D\}$ and $D_y=\{x\in\mathbb{R}\mid (x,y)\in D\}$ belongs to $\mathcal{B}(\mathbb{R})$.
For (2) you just evaluate the inner integrals first: For a fixed $y\in [0,1]$ we have that $\chi_D(x,y)=\chi_{D_y}(x)=1$ if and only if $x=y$ and zero otherwise. Therefore,
$$
\int_{[0,1]}\chi_D(x,y)\,m(\mathrm dx)=\int_{[0,1]}\chi_{D_{y}}(x)\,m(\mathrm dx)=m(\{y\})=0,
$$
for all $y\in [0,1]$.
For the right-hand side we have that for a fixed $x\in [0,1]$:
$$
\int_{[0,1]}\chi_D(x,y)\,\nu(\mathrm dy)=\int_{[0,1]}\chi_{D_{x}}(y)\,\nu(\mathrm dy)=\nu(\{x\})=1.
$$
Is this a contradiction to Tonelli/Fubini's theorem? (This is probably the key point of the exercise).
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
Since we can only take finite partitions, a natural choice would be to consider the partition $P_m$ of $\mathbb{N}$ that consists of $ \{0\}, \{1\}, \{2\}, \{3\}, \dots, \{m\}$ and $\{m+1,m+2,\dots\}$.
Then the singleton sets have counting measure $1$ while the last set has counting measure $\infty$, so that $$\mathcal{L}(f, P_m) = \sum_{k=0}^m 1\cdot b_k + \infty\cdot \inf_{k>m} b_k.$$ (with the usual convention that $\infty \cdot 0 = 0$).
There are now two cases: