The answer to the local question is no.

Consider on $\mathbb RP(2)$ the vector field defined on an affine chart $(x,y)\in \mathbb R^2$ by $v=x \partial_x + y \partial y$. The induced flow acts on $\mathbb R^2$ by homotheties, an therefore fixes the line at infinity $ell$. Hence $v$ extends to a vector field on $\mathbb RP(2)$ and contains $\ell$ in its zero set. Writing down explicitly how the vector field looks like at a neighborhood of point at the line at infinity, one sees that it vanishes in first order at it. The natural thing to do is to find a function vanishing at the line at infinity at first order and divide $v$ by it. But this cannot be done. A neighborhood of $\ell$ is isomorphic to the Moëbius band, with $\ell$ being the central circle. A function $f$ vanishing exactly on $\ell$ and only up to first order (i.e. $\ell$ is not contained in the critical set of $f$) would allow us to distinguish, through its sign, up and down the central circle of the Moëbius band.

A version of the local question is true.

The zeros of an analytic vector field, if not isolated, must contain an analytic curve. Locally this curve is defined by the vanishing of an analytic function. Dividing by the equation of the curve one gets another vector field with isolated zeros.

Anyway, I don't think that these issues cause any kind of problem.

In general a singular analytic foliation on a compact surface $S$ is not defined by a global analytic vector field. Usually one covers $S$ with open subsets where the foliation is defined by vector fields with isolated zeros, and ask that on non-empty intersections the vector fields differ by the multiplication with a nowhere vanishing analytic function.

Disclaimer: The discussion below is about vector fields on $\mathbb RP(2)$, a non-orientable surface, and they do not answer the OP's questions.

The answer to the local question is no.

Consider on $\mathbb RP(2)$ the vector field defined on an affine chart $(x,y)\in \mathbb R^2$ by $v=x \partial_x + y \partial y$. The induced flow acts on $\mathbb R^2$ by homotheties, an therefore fixes the line at infinity $ell$. Hence $v$ extends to a vector field on $\mathbb RP(2)$ and contains $\ell$ in its zero set. Writing down explicitly how the vector field looks like at a neighborhood of point at the line at infinity, one sees that it vanishes in first order at it. The natural thing to do is to find a function vanishing at the line at infinity at first order and divide $v$ by it. But this cannot be done. A neighborhood of $\ell$ is isomorphic to the Moëbius band, with $\ell$ being the central circle. A function $f$ vanishing exactly on $\ell$ and only up to first order (i.e. $\ell$ is not contained in the critical set of $f$) would allow us to distinguish, through its sign, up and down the central circle of the Moëbius band.

A version of the local question is true.

The zeros of an analytic vector field, if not isolated, must contain an analytic curve. Locally this curve is defined by the vanishing of an analytic function. Dividing by the equation of the curve one gets another vector field with isolated zeros.

Anyway, I don't think that these issues cause any kind of problem.

In general a singular analytic foliation on a compact surface $S$ is not defined by a global analytic vector field. Usually one covers $S$ with open subsets where the foliation is defined by vector fields with isolated zeros, and ask that on non-empty intersections the vector fields differ by the multiplication with a nowhere vanishing analytic function.