It is tacitly understood that a differentiable and regular meridian curve $\mu:\quad t\mapsto \bigl(r(t),z(t)\bigr)\qquad(a with $r(t)>0$ is given. Then the map $f:\ {\mathbb R}\times\ ]a,b[\ \to{\mathbb R}^3,\qquad (\phi,t)\mapsto\bigl(r(t)\cos\phi,r(t)\sin\phi, z(t)\bigr)$ has a differential $df$ of rank $2$ for all $(t,\phi)$, as $f_t\times f_\phi\ne 0$ everywhere. It follows that $f$ is an immersion, but it produces each piece of the intended surface of revolution $S$ an infinite number of times.
In order to obtain $S$ just once one could proceed to a quotient space ($\phi$ runs over ${\mathbb R}/(2\pi)$ instead of ${\mathbb R}$), or one covers $S$ with a finite number of "patches" that may overlap, but are related via "local coordinate transformations".
With only two patches there is the difficulty that they have a disconnected overlap, and two different local coordinate transformations apply in the two parts. Therefore I would propose to set up three patches as follows: $p_1:\quad \bigl]{-{\pi\over2}},{\pi\over2}\bigr[\ \times\ ]a,b[\ \to{\mathbb R}^3,\qquad (\phi,t)\mapsto\bigl(r(t)\cos\phi,r(t)\sin\phi, z(t)\bigr)\ ,$ $p_2:\quad \bigl]{{\pi\over4}},{5\pi\over4}\bigr[\ \times\ ]a,b[\ \to{\mathbb R}^3,\qquad (\psi,t)\mapsto\bigl(r(t)\cos\psi,r(t)\sin\psi, z(t)\bigr)\ ,$ $p_3:\quad \bigl]{{3\pi\over4}},{7\pi\over4}\bigr[\ \times\ ]a,b[\ \to{\mathbb R}^3,\qquad (\omega,t)\mapsto\bigl(r(t)\cos\omega,r(t)\sin \omega, z(t)\bigr)\ .$ In this way we have the three local coordinate transformations (only the angle variable is affected) $p_1^{-1}\circ p_2: \quad \psi\mapsto \phi=\psi\qquad \bigl({\pi\over4}<\psi<{\pi\over2}\bigr)\ ,$ $p_2^{-1}\circ p_3: \quad \omega\mapsto \psi=\omega\qquad \bigl({3\pi\over4}<\omega<{5\pi\over4}\bigr)\ ,$ $p_3^{-1}\circ p_1: \quad \phi\mapsto \omega=\phi+2\pi\qquad \bigl({-{\pi\over2}}<\phi<{-{\pi\over4}}\bigr)\ .$ Please note that everything we have written up here should be intuitively obvious and only served as an exercise in applying the general theory in a familiar case.