Covariance function of Brownian motion and the second derivative operator

I'm not sure what "deeper reason" you are aiming at, but for one thing, there is no surprise here. The restriction of $W_n$ of $W$ to $(t_1,\dots,t_n)$ is a Gaussian vector whose density form is given by

$$ (\Sigma^{-1}_n v;v)=v^2_1\cdot t_1+\sum_{i=2}^n (v_i-v_{i-1})^2(t_i-t_{i-1})=(\nabla_n v;\nabla_n v)=(\nabla_n^*\nabla_n v;v), $$ where $(\cdot,\cdot)$ is the usual dot product and $\nabla_n$ and $\nabla^*_n$ are the finite difference operators: $$ (\nabla_n v)_i = (v_i-v_{i-1})(t_i-t_{i-1})^\frac12;\quad (\nabla^*_n v)_i = (v_{i+1}-v_{i})(t_{i+1}-t_{i})^\frac12. $$ The matrices of these operators are bidiagonal, so their product is always tridiagonal. Actually, it is easy to see that for any Gaussian process with Markov property (e. g. the Ornstein-Uhlenbeck), the inverse covariance matrix will be tridiagonal, and so it can be written as a linear combination of finite difference operators up to second order (generally, with variable coefficients).

Generally, the infinitesimal generator of a stochastic process (e. g. Ito diffusion) is not uniquely determined by its mean and covaraince: if $\mathbb{E}X_t\equiv 0$, then $X_t$ and $-X_t$ have the same mean and covariance, but, of course, need not have the same distribution.