Let me first try to make sense of your question. First of all, the notion of a differential form is meaningless for general metric spaces $(X,d)$. However, one can still talk about Borel measures $\mu$ on the topological space $X$ (topologized using the metric $d$). The Borel condition still leaves too much freedom since we did not use the metric $d$ (only the topology). The most common condition these days is the one of a metric measure space which ties nicely $d$ and $\mu$ and allows one to do quite a bit of analysis on $(X,d,\mu)$ similarly to the analysis on the Euclidean $n$-space $E^n$. (The literature on this subject is quite substantial, just google "metric measure space".)
Definition. A triple $(X,d,\mu)$ (where $d$ is a metric on $X$ and $\mu$ is a Borel measure on $X$) is called a metric measure space if the measure $\mu$ is doubling with respect to $d$, i.e. there exists a constant $D<\infty$ such that for every $a\in X, r>0$, we have
$$
\mu(B(a, 2r))\le D \mu(B(a,r)),
$$
where $B(a,R)$ denotes the closed ball of radius $R$ centered at $a$.
Example. Every closed subset of $E^n$ (equipped with the restriction of the Euclidean metric) is a complete doubling metric space.
The basic existence result for doubling measures $\mu$, proven in
"Every complete doubling metric space
carries a doubling measure", by J. Luukainen, E. Saksman, Proc. Amer. Math. Soc. 126 (1998) p. 531–534, is the following:
Theorem. A complete metric space $(X,d)$ carries a doubling measure $\mu$ if and only if the metric $d$ is doubling, i.e. there exists a constant $C$ such that every ball of radius $2r$ in $X$ is covered by at most $C$ balls of radius $r$.
The proof of this theorem is not long but by no means trivial.
Let $L=\inf_{m\in M}(f(m))$ and $A=\sup_{m\in M}(f(m))$. By assumption $L, A$ exist and are positive. We define two metric structures, $d,d': M\times M\to \mathbb{R}^*$, $$d(x,y)=\inf \bigg\{\int_{0}^1( g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt: \gamma \text{ is a piecewise differentiable curve from } a \text{ to } b \bigg\}$$
And $$d'(x,y)=\inf \bigg\{\int_{0}^1( f^2g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt: \gamma \text{ is a piecewise differentiable curve from } a \text{ to } b \bigg\}$$
Since $(M,g)$ is geodesically complete, $d$ makes $M$ into a complete metric space by the Hopf-Rinow theorem. It suffices to show that the metrics $d$ and $d'$ are strongly equivalent, i.e. for all $x,y$ in $M$, there exist $\alpha,\beta\in \mathbb{R}$ such that $\alpha d(x,y)\leq d'(x,y)\leq \beta d(x,y)$.
We can see that for any piecewise differentiable curve $\gamma:[0,1]\to M$ such that $\gamma(0)=x$ and $\gamma(1)=y$ we have $L(g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}\leq(f^2g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}\leq A(g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}$ implying that $$L\int_{0}^1(g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt\leq \int_{0}^1 (f^2g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt\leq A\int_{0}^1(g_{ij}\dot{\gamma} ^i\dot{\gamma} ^j)^{1/2}dt$$ and hence $$Ld(x,y)\leq d'(x,y)\leq Ad(x,y)$$ So $d$ and $d'$ are strongly equivalent. Because $(M,d)$ is a complete metric space and $(M,d')$ is strongly equivalent to $(M,d)$, then it is also a complete metric space and $(M,f^2g)$ is geodesically complete.
Best Answer
ADDED: More detailed explanation:
This is a metric on a circular cylinder of varying radius. $t$ is a parameterization of the central axis, and for each $t$, $f(t)$ is the radius of the circle at that point on the axis.
If $f$ is odd, then $f(0)= 0$ and the cylinder "pinches" down to a point as $t \rightarrow 0$. In general this results in a cone with a singular point at $t = 0$. A natural question is when is the surface in fact smooth with a smooth metric for $t$ near $0$.
The assumptions that $f'(0) = 1$ and $f$ is odd imply that the surface and metric are smooth at the origin. To see this a little more easily, it's better to change the variable names from $t$ and $x$ to $r$ and $\theta$.
If $x = r\cos\theta$ and $y = r\sin\theta$, then $$ dx = dr\,\cos\theta - d\theta\,r\sin\theta\text{ and }dy = dr\,\sin\theta + d\theta\,r\cos\theta $$ Solving for $dr$ and $d\theta$, we get \begin{align*} dr &= \frac{x\,dx + y\,dy}{r}\\ d\theta &= \frac{-y\,dx + x\,dy}{r^2}. \end{align*} Now assume that $f$ can be written as $f(r) = r\phi(r)$. Then using the formulas above and doing some algebra, you get \begin{align*} g &= dr^2 + f^2\,d\theta^2\\ &= \frac{(x^2 + y^2\phi^2)\,dx^2 + 2xy(1-\phi)\,dx\,dy + (y^2 + x^2\phi^2)\,dy^2}{r^2} \end{align*} If $\phi$ is even and $\phi(0) = 1$, then it is easy to check that this metric is smooth with respect to $(x,y)$ including at the origin. These conditions on $\phi$ are equivalent to $f'(0) = 1$ and $f$ odd.
The standard examples are flat Euclidean space, where $f = r$, the sphere of radius $R$, where $$ f = R\sin\frac{r}{R}, $$ and hyperbolic space, where $$ f = \sinh r $$