First, the classical correspondence, explaining why the SLD should be present. The Fisher information is the expectation value of the score, where the score is the logarithmic derivative of the probability distribution: $$ L_{\mathrm{classical}}=\frac{\partial [\ln p(\pmb{x}|\theta)]}{\partial \theta}, $$ which leads to the relation $$ \frac{\partial [ p(\pmb{x}|\theta)]}{\partial \theta}= p(\pmb{x}|\theta)L_{\mathrm{classical}}=L_{\mathrm{classical}} \,p(\pmb{x}|\theta). $$ This clearly mirrors the expression for the SLD $L$, where the anticommutator is required to resolve the operator ordering ambiguity - it justifies the origin of the operator, but as of yet does not lend us too much geometric insight.
Now for some geometry. This is mostly taken from the older geometry paper and the recent geometrical review (the latter is available on arXiv but I'm not sure about the former). Various probability distributions $p(\pmb{x}|\theta)$ for a given parameter $\theta$ form a probability simplex and we are looking for a metric to know the distance between two distributions. A good candidate is the Fisher-Rao metric (here written for discrete variables) $$ ds^2=\sum_{jk}g_{jk} dp^j dp^k=\sum_j\frac{(d p^j)^2}{p_j}=\sum_j p_j (d\ln p^j)^2. $$ The metric makes sense for classical variables with expectation values $$\langle A\rangle=\sum_j A_j p^j,\quad \langle AB\rangle=\sum_{jk}A_j B_k g^{jk}=\sum_j A_j B_j p^j$$ because those expectation values constrain the metric tensor to obey $$g_{jk}=\frac{\delta_{jk}}{p^j}.$$ Displacement along a line segment $d\theta$ then leads us to the classical Fisher information $$ F=\frac{ds^2}{d\theta^2}=\sum_j p_j\left(\frac{\partial \ln p^j}{\partial \theta}\right)^2. $$ This already looks very similar to the metric induced by the Bures distance, if we recall the correspondence between the score and the SLD.
When we do the same thing with operators and probability distributions found via the Born rule, we see the need to somehow divide by the probability distribution. This is achieved via the superoperator $$ \mathcal{R}_{\rho}(O)=\frac{\left\{\rho,O\right\}}{2}, $$ with which the correlation between two observables can be written using $$ \left\langle \frac{\left\{A,B\right\}}{2}\right\rangle=\mathrm{Tr}\left[A \mathcal{R}_\rho (B)\right]=\mathrm{Tr}\left[B \mathcal{R}_\rho (A)\right]. $$ In this expression, $R_\rho(\cdot)$ plays the role of the metric tensor $g^{jk}$. To find the operator playing the role of the metric tensor with lowered indices, we want the inverse of this operator $\mathcal{L}_\rho(\cdot)=\mathcal{R}^{-1}_\rho(\cdot)$ (because $\sum_{k}g_{jk}g^{kl}=\delta_{kl}$), which must obey $$ \mathcal{L}_\rho[\mathcal{R}_\rho(B)]=B. $$ We can then use this to construct a scalar product between quantum states as $$ \langle\sigma_1,\sigma_2\rangle=\mathrm{Tr}\left[\sigma_1\mathcal{L}_\rho(\sigma_2)\right] $$ and with that define the infinitesimal distance $$ ds^2=\mathrm{Tr}\left[d\rho\mathcal{L}_\rho(d\rho)\right]. $$ The quantum Fisher information then takes the form $$ Q=\frac{ds^2}{d\theta^2}=\mathrm{Tr}\left[\frac{\partial\rho}{\partial \theta}\mathcal{L}_\rho\left(\frac{\partial \rho}{\partial \theta}\right)\right]. $$
The only remaining ingredient is to identify the SLD as $$L=\mathcal{L}_\rho\left(\frac{\partial \rho}{\partial \theta}\right),$$ which can be proven by expanding all of the operators in the eigenbasis of $\rho$. We then see how the SLD is related to the classical score, the distance between classical probabability distributions, and the distance between quantum states.
TL;DR: We're looking for an inner product between $\partial\rho/\partial \theta^\mu$ and $\partial\rho/\partial \theta^\nu$, and the metric for the inner product is related to the SLD $L_\nu=\mathcal{L}_\rho(\partial \rho/\partial \theta^\nu)$ because this metric makes classical expectation values like $\langle A\rangle$ and $\langle AB\rangle$ agree with each other.
Extra:
The inverse operator satisfies $$ \mathrm{Tr}(AB)=\mathrm{Re}\left\{\mathrm{Tr}\left[\rho A\mathcal{L}_\rho(B)\right]\right\}. $$