Setup. Let $(M,g)$ be a Riemannian manifold of dimension $2$, let $m\in M$ and let $U$ be an open neighborhood of $0$ in $T_mM$ such that $\exp_m\colon U\rightarrow M$ is an embedding and let $h:={\exp_m}^*g$.
According to Gauss' lemma, in polar coordinates $[r,\theta]$ on $U\setminus\{0\}$, one has the following decomposition:
$$h=\mathrm{d}r^2+f(r,\theta)^2\mathrm{d}\theta^2.$$
Let $u$ and $v$ be two vectors of $T_mM$ which are orthogonal for $h_0$ and assume that polar coordinates have been chosen such that the angle associated to $u$ is $0$.
Question. Let $c\colon r\mapsto ru$, let $r\mapsto V(r)$ be the parallel transport of $v$ along $c$ and let $\cdot'$ be the connection induced on $c^*TM$ by $\nabla$ the Levi-Civita connection of $M$, then: $$J(r):=f(r,0)V(r)$$ is the Jacobi field along $c$ with initial values $J(0)=0$ and $J'(0)=v$.
My attempts.
I have to prove that $J$ satisfies the following second order linear differential equation: $$J''-R(\dot{c},J)\dot{c}=0.$$ The first term is easy to compute, as $V$ is parallel along $c$, one has: $$J''(r)=\frac{\partial^2f}{\partial r^2}(r,0)V(r).$$ However, I am having a hard time with the curvature term, I am aware that by definition, one has: $$R(\dot{c},J)\dot{c}=\nabla_{\dot{c}}\nabla_J\dot{c}-\underbrace{\nabla_{J}\nabla_{\dot{c}}{\dot{c}}}_{=0}-\nabla_{[\dot{c},J]}\dot{c}$$ and the middle term is vanishing since $c$ is a geodesic path and here I am stuck.
An other strategy I have in mind is finding a smooth map $\ell\colon I\times]-\varepsilon,\varepsilon[\rightarrow M$ such that: $$J(r)=\frac{\partial\ell}{\partial s}(r,0)$$ and for all $s$, $t\mapsto\ell(t,s)$ is a geodesic of $M$. I have not dig in depth this approach yet.
Another idea could be to prove that one has $J(r)=T_{ru}\exp_m(v)$, but it does not seem too doable.
Any enlightenment will be greatly appreciated.
Your other strategy (number 2) is a good one.
Note that using your polar coordinates, the path$$r\mapsto(r,\theta_0)$$is a geodesic ray for every fixed $\theta_0$. The variation of this family of geodesics at $\theta_0=0$ is simply $\partial/\partial\theta,$ and so, $\partial/\partial\theta$ is indeed a Jacobi field along the geodesic ray in question.
On the other hand, we can compute $V(r)$. As parallel translation with respect to the Levi-Civita connection is an isometry, the tangent vector $V(r)$ is orthogonal to $\partial/\partial r$ at every point, and $|V(r)|$ is constant in $r$. As $$\left|\frac{\partial}{\partial\theta}\right|=f,$$we conclude that $$V(r)=\frac{|v|}{f}\frac{\partial}{\partial\theta},$$and so,$$J(r)=|v|\frac{\partial}{\partial\theta}.$$
Remark: The above calculations hold at every point where $r\neq0$ and are meaningless at the origin. This only means that you will need some continuity argument for a complete solution.