The LRN theorem lets us decompose one measure with respect to another -- one part singular and one part absolutely continuous. If $\nu \ll \mu$, then there can be no singular part.
This is because $\nu \ll \mu$ says that if $\mu(E) = 0$, then $\nu(E) = 0$ too. But if $\nu = \lambda + \rho$ with $\lambda \perp \mu$, then $X = E \sqcup F$ where $F$ is $\mu$-null and $E$ is $\lambda$-null, but then $F$ is $\mu$-null, $F$ must also be $\nu$-null because in this case, $\nu \ll \mu$, so $F$ must also be $\lambda$-null. But then $X$ is $\lambda$-null, so that $\lambda \equiv 0$.
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
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).