I am surprised by the fact that $\mathrm{SU}(1, 1)$ group is isomorphic to $\mathrm{SL}(2, \mathbb{R})$, but $\mathrm{SU}(2)$ is not isomorphic to $\mathrm{SL}(2, \mathbb{R})$.
The first statement is easy to prove. An element of $\mathrm{SU}(1, 1)$ is
$$g=\left[\begin{array}{cc} \alpha & \beta \\ \beta^* & \alpha^* \end{array}\right] =\left[\begin{array}{cc} x+iy & z+id\\ z-id & x-iy \end{array}\right] \quad , |\alpha|^2-|\beta|^2=1.$$
The mapping to $p=\left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \in \mathrm{SL}(2, \mathbb{R})$, $(ad-bc=1) $ is $$a=x-d, \quad b=z-y, \quad c=y+z, \quad d=x+d.$$ The second fact is well known and discussed here SU(2) and SL(2,R) are not isomorphic
Is there any intuitive way to understand why $\mathrm{SU}(1, 1)$ and $\mathrm{SU}(2)$ are so different in such respect?
The group $SU(2)$ is compact, whereas $SL(2,\Bbb R)$ isn't. That's enough to make a huge difference. And $SU(2)$ is compact because the condition that defines it ($|\alpha|^2+|\beta|^2=1$) assures that it is a closed and bounded subset of $\Bbb C^{2\times2}$. Clearly, $SU(1,1)$ is unbounded.