A proof of this can be extracted from Steinberg's paper "Générateurs, relations et revêtements de groupes algébriques". If Ive understood it correctly, he shows that a Cartan subgroup of the universal central extension of $SL_2(K)$ (for a field $K$ with at least 4 elements) is generated by elements $h(t)$ for $t$ a nonzero element of the field, that in particular satisfy $h(tu^2)=h(t)h(u^2)$, and the center of the universal central extension is the kernel of the map from the Cartan subgroup to $K^*/\pm 1$ taking $h(t)$ to $t$. In the case when $K$ is the reals, this kernel is generated by $h(-1)$ as every element is a square or a square times $-1$. So the center of the universal central extension is generated by $h(-1)$ and in particular is cyclic. The center is known be be at least $Z$, so the center of the universal central extension is exactly $Z$, and the universal central extension is therefore the same as the universal cover.

(This is incorrect: see below)

A proof of this can be extracted from Steinberg's paper "Générateurs, relations et revêtements de groupes algébriques". If Ive understood it correctly, he shows that a Cartan subgroup of the universal central extension of $SL_2(K)$ (for a field $K$ with at least 4 elements) is generated by elements $h(t)$ for $t$ a nonzero element of the field, that in particular satisfy $h(tu^2)=h(t)h(u^2)$, and the center of the universal central extension is the kernel of the map from the Cartan subgroup to $K^*/\pm 1$ taking $h(t)$ to $t$. In the case when $K$ is the reals, this kernel is generated by $h(-1)$ as every element is a square or a square times $-1$. So the center of the universal central extension is generated by $h(-1)$ and in particular is cyclic. The center is known be be at least $Z$, so the center of the universal central extension is exactly $Z$, and the universal central extension is therefore the same as the universal cover.

Added later: this answer is wrong. I overlooked that the proof of $h(tu^2)=h(t)h(u^2)$ in Steinberg's paper requires that $t$ and $u$ generate a cyclic subgroup of $K^*$, which holds in the finite field case he was considering but not over the reals.

A proof of this can be extracted from Steinberg's paper "Générateurs, relations et revêtements de groupes algébriques". If Ive understood it correctly, he shows that a Cartan subgroup of the universal cover of $SL_2(K)$ (for a field $K$ with at least 4 elements) is generated by elements $h(t)$ for $t$ a nonzero element of the field, that in particular satisfy $h(tu^2)=h(t)h(u^2)$, and the center of the universal central extension is the kernel of the map from the Cartan subgroup to $K^*$ taking $h(t)$ to $t$. In the case when $K$ is the reals, this kernel is generated by $h(-1)$ as every element is a square or a square times $-1$. So the center of the universal central extension is generated by $h(-1)$ and in particular is cyclic. The center is known be be at least $Z$, so the center of the universal central extension is exactly $Z$.

A proof of this can be extracted from Steinberg's paper "Générateurs, relations et revêtements de groupes algébriques". If Ive understood it correctly, he shows that a Cartan subgroup of the universal central extension of $SL_2(K)$ (for a field $K$ with at least 4 elements) is generated by elements $h(t)$ for $t$ a nonzero element of the field, that in particular satisfy $h(tu^2)=h(t)h(u^2)$, and the center of the universal central extension is the kernel of the map from the Cartan subgroup to $K^*/\pm 1$ taking $h(t)$ to $t$. In the case when $K$ is the reals, this kernel is generated by $h(-1)$ as every element is a square or a square times $-1$. So the center of the universal central extension is generated by $h(-1)$ and in particular is cyclic. The center is known be be at least $Z$, so the center of the universal central extension is exactly $Z$, and the universal central extension is therefore the same as the universal cover.