If I have a curve parametrized in the form $(f(v),0,g(v)),f>0, v\in(a,b)$, rotating this curve about the $z$-axis. I get a surface of revolution $S$ and I know it is a regular surface as it can be parametrized by two local charts $X_1(u,v)=(\cos u f(v),\sin u f(v),g(v)), u\in(0,2\pi)$ and $X_2(u,v)=(\cos u f(v),\sin u f(v),g(v)), u\in(-\frac{\pi}{2},\frac{\pi}{2})$
Then I know that if I have a local chart, then there exists at least one unitary normal vector field on $X_1(U)$, i.e there exists $N: X_1(U)\rightarrow\mathbb{R^3}$ The problem is, how can I make sure there exists a unitary normal vector field on the whole surface, i.e $N:S\rightarrow\mathbb{R^3}$ which proves that this surface is orientable?
You're 90% of the way there.
Take point $P$ in the overlap of the two charts, a point $P$ with coordinates $(u, v)$.
Using $X_1$, we get two tangent vectors, \begin{align} X_{1u}(u, v) &= (-f(v) \sin u, f(v) \cos u, 0) \\ X_{1v}(u, v) &= (f'(v) \cos u, f'(v) \sin u, g'(v)) \\ \end{align} and computing their cross product, we get a normal vector \begin{align} n_1(u, v) &= ( f(v)g'(v) \cos u, f(v) g'(v) \sin u, -f(v) f'(v)) \end{align} from which we can (since we'll be normalizing it anyhow) remove a factor of $f(v)$ (because $f$ is everywhere positive!), so I'll rewrite \begin{align} n_1(u, v) &= ( g'(v) \cos u, g'(v) \sin u, -f'(v)) \end{align} The squared length of this vector is $g'(v)^2 + f'(v)^2$, so the unit normal determined by the first parameterization at location $(u, v)$ is \begin{align} N_1(u, v) &= \frac{1}{\sqrt{g'(v)^2 + f(v)^2}}( g'(v) \cos u, g'(v) \sin u, -f'(v)). \end{align}
If we do exactly the same computation for $N_2$, we end up with exactly the same formula. This shows us that at points with $0 < u < \frac{\pi}{2}$, the two computed unit-normal fields are identical.
What about points with $-\frac{\pi}2 < u < 0$ (for $X_2$), which correspond to points $3\pi/2 < u' < 2\pi$ for $X_1$, subject to $$ X_2(u) = X_1(u + 2\pi)? $$ If we let $u' = u + 2\pi$, then we can compare $N_2(u)$ with $N_1(u') = N_1(u + 2\pi)$. But the formula for $N_1$ (and for $N_2$) is periodic of period $2\pi$, so we find that $N_1(u) = N_2(u')$, as required.
For every point on the surface with coordinates $(u, v)$ (in either coordinate system!), we find that the formula for $N_1(u, v)$ given above gives a unit vector field at the surface-point corresponding to $(u, v)$, and does so in a way that varies continuously in the coordinates, as required for a normal field.