Because of Iwasawa Decomposition(link), I know that $SL(2,\mathbb{R})$ is also decomposable by $K,A,$ and $N$, where $K$ is $SO(2)$, $A$ is the set of diagonal matrices with $(x,\frac{1}{x})$, and $N$ is the group of matrices $((1,x),(0,x))$.
However, when I read "Ergodic theory, with a view towards number theory," which is written by Manfred Einsiedler, it states that on the page 284,
Now the subgroup U = $\left\{\begin{pmatrix}1 & b \\ 0 & 1 \end{pmatrix}: b \in \mathbb{R}\right\} \leq SL_{2}{(\mathbb{R})}$ together with
$w = \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}$ generates $SL_{2}{(\mathbb{R})}$, since $wUw^{-1} = \left\{\begin{pmatrix}1 & 0 \\ -b & 1 \end{pmatrix}: b \in \mathbb{R} \right\}$.
But, I don't know how to this $w$ and $U$ generates the whole $SL_{2}(\mathbb{R})$. How can I show that they are generator of $SL_{2}(\mathbb{R})$?
This is 4 years late, but here's an answer for others. All lower and upper triangular matrices mentioned below are to have 1s on the diagonal, they are taken from the given set of generators.
First, we look at the product $$\begin{bmatrix}1&0\\\alpha&1\end{bmatrix} \begin{bmatrix}x&0\\0&1/x\end{bmatrix} \begin{bmatrix}1&\beta\\0&1\end{bmatrix} = \begin{bmatrix} x&\beta x\\\alpha x& \alpha\beta x+1/x\end{bmatrix}$$
Given $\begin{bmatrix}a&b\\c&d\end{bmatrix}\in SL_2(\mathbb{R})$, when $a\neq 0$, we can solve for $\alpha, x, \beta$. Since we have lower and upper triangular matrices, it then suffices to show that the generators in question also generate the diagonal matrices. We will come to the $a = 0$ case later.
Now we look at a product of the form LULU: $$\begin{bmatrix} 1&0\\\alpha&1\end{bmatrix} \begin{bmatrix} 1&\beta\\0&1\end{bmatrix} \begin{bmatrix} 1&0\\\gamma&1\end{bmatrix} \begin{bmatrix} 1&\delta\\0&1\end{bmatrix}= \begin{bmatrix} 1+\beta\gamma&\delta(1+\beta\gamma)+\beta\\ \alpha(1+\beta\gamma)+\gamma&\alpha\delta(1+\beta\gamma)+\alpha\beta+\gamma\delta+1 \end{bmatrix}$$
Assuming $1+\beta\gamma\neq 0$, to get the anti-diagonal terms to be $0$, we take $$\alpha = \frac{-\gamma}{1+\beta\gamma}, \delta = \frac{-\beta}{1+\beta\gamma}$$ then the bottom right term becomes $$\frac{\beta\gamma}{1+\beta\gamma}+\frac{-2\beta\gamma}{1+\beta\gamma}+1 = \frac{1}{1+\beta\gamma}$$ as required.
Therefore, a matrix of the form $\begin{bmatrix}x&0\\0&1/x\end{bmatrix}$ is generated by lower and upper diagonal matrices with ones on the diagonal for suitable choice of $\beta, \gamma$.
When $1+\beta\gamma = 0$, the above product becomes $$\begin{bmatrix} 0&\beta\\ \gamma& 1+\alpha\beta+\gamma\delta \end{bmatrix}$$
We can take $\delta = 0$ and make suitable choices so that the matrix becomes $\begin{bmatrix} 0&b\\c&d\end{bmatrix}\in SL_2(\mathbb{R})$.
So, any matrix in $SL_2(\mathbb{R})$ is of the form LUL or LULU where the matrices have 1s on the diagonal. The same argument holds good over other fields as well.