The proof of this theorem in Folland provides quite some insight into the construction of
$$
\mathrm d\nu = f\mathrm d\mu + \mathrm d\lambda
$$
where $\lambda\perp \mu$. First of all, the function $f$ is constructed using the class
$$
\mathscr F = \left\{f:X\to [0,\infty]:\int_E f\mathrm d\mu\leq \nu(E)\text{ for all }E\in \mathscr M\right\}.
$$
Secondly, one defines $a := \sup\{\int _X f\mathrm d\mu|f\in \mathscr F\}$ and as it follows from the definition of $a$, there exists a sequence $(f_n)_{n\in \Bbb N} \subset \mathscr F$ such that
$$
\int_X f_n\mathrm d\mu \to a
$$
The theorem claims that $f := \sup_nf_n$ is indeed such a function that
$$
\lambda := \int f\mathrm d\mu - \nu\perp\mu.
$$
As a result, the existence of the function $f$ is proved by constructing it. However, I often saw the statement that "there is no known constructive proof of Radon-Nikodym Theorem", which I guess means that in practice in general it's hard to get the explicit expression of $f$ (provided one can define what does the "explicit expression" mean).
There is no need for the Radon-Nikodým theorem.
By the very definition of the Radon-Nikodým derivative, we are looking for a function $g: (0,\infty) \to [0,\infty)$ which is measurable with respect to $\mathcal{A}$ and satisfies $$\lambda(E) = \int_E g \, dm, \qquad E \in \mathcal{A}. \tag{1}$$
Note that $g(x) := f(x) :=2x^2$ is not measurable with respect to $\mathcal{A}$ and therefore we cannot simply choose $g=f$.
If we define
$$E_n := \begin{cases} \bigg( \frac{1}{n+1}, \frac{1}{n} \bigg], & n \in \mathbb{N}, \\ (1,\infty), & n = 0 \end{cases}$$
then $\mathcal{A} = \sigma(E_n; n \in \mathbb{N}_0)$. Since the intervals $E_n$, $n \in \mathbb{N}_0$, are disjoint and cover $(0,\infty)$, equation $(1)$ is equivalent to
$$\lambda (E_n) = \int_{E_n} g \, dm \qquad \text{for all} \, \, n \in \mathbb{N}_0. \tag{2}$$
Moreover, any $\mathcal{A}$-measurable function $g$ is of the form
$$g(x) = \sum_{n \in \mathbb{N}_0} c_n 1_{E_n}(x) \tag{3}$$
for constants $c_n \in \mathbb{R}$. The only thing which we have to do is to choose the constants $c_n \geq 0$ such that $(2)$ holds. To this end, we plug our candidate $(3)$ into $(2)$ and find
$$\lambda(E_n) \stackrel{!}{=} \int_{E_n} g \, dm \stackrel{(3)}{=} c_n \int_{E_n} \, dm= c_n m(E_n) = c_n \left( \frac{1}{n}-\frac{1}{n+1} \right)$$
which implies
$$c_n = \lambda(E_n) n (n+1)$$
for all $n \in \mathbb{N}$. $\lambda(E_n)$ can be calculated explicitly using the very definition of $\lambda$; I leaves this to you. For $n=0$ we get $$\lambda(E_0) = \infty \stackrel{!}{=} c_0 m(E_0) = c_0 \infty,$$ i.e. we can choose $c_0 := 1$. Hence,
$$g(x) = 1_{E_0}(x)+ \sum_{n \geq 1} \lambda(E_n) n (n+1) 1_{E_n}(x)$$
is a non-negative $\mathcal{A}$-measurable function which satisfies $(2)$ (hence, $(1)$), i.e.
$$g = \frac{d\lambda}{dm}.$$
Best Answer
According to Lebesgue's decomposition theorem, the measure $\mu$ decomposes as $$\mu = \rho + \nu,$$ where
$\rho$ is absolutely continuous with respect to the Lebesgue measure $m$;
$\nu$ and $m$ are singular with respect to each other.
Furthermore, this decomposition is unique.
This uniqueness of decomposition is key. It means that a possible way of solving this problem is to guess what $\rho$ and $\nu$ might be, based on intuition, then verify our the guesses obey the above properties.
As you say, it's not hard to come up with the right intuition here. To paraphrase your description:
The "continuous" part of the measure is supported on $(1, \infty)$. By elementary calculus, we would expect that this continuous part is given by the $\rho(E) = \int_{\mathbb R} \pi .1_{E \cap (1,\infty)} \ dm$.
The "discrete" part of the measure is supported on $\{ 1 \}$. It is simply $\nu(E) = \pi $ if $1 \in E$, or $\nu(E) = 0$ if $1 \notin E$.
So what needs to be done to formalise this?
We need to show that my $\rho$ really is absolutely continuous w.r.t. $m$. This should be obvious, given the definition of $\rho$. It should also be obvious what the Radon-Nikodym derivative $d\rho / dm$ is.
We need to show that $\nu$ and $m$ are singular. This shouldn't be too difficult either.
We need to show that $\mu = \rho + \nu$. This can be done by switching to polar coordinates using the Jacobian change of variables formula...