Let $M$ be a manifold, $g$ a Riemannian metric, $\nabla$ be an affine connection on $M$, and $\nabla^{*}$ the unique dual affine connection of $\nabla$ on $M$, i.e. for all vector fields $X,Y,Z$ on $M$, we have $$Zg(X,Y)=g(\nabla_{Z} X,Y)+g(X,\nabla_Z^{*} Y)$$ If both $\nabla$ and $\nabla^{*}$ are torsion-free, then we call $(M,g,\nabla,\nabla^{*})$ a statistical manifold.
On the following Wikipedia article, it is stated that if you have a probability space $(M,\mathcal{F},\mu)$, where $M$ is an orientable manifold, then you can induce an infinite-dimensional statistical manifold $S(M)$ defined to be the space of all probability measures on $M$ (with $\mathcal{F}$ held fixed).
My question is why should $M$ be orientable to be able to induce this statistical manifold?
I can see why one can induce such a statistical manifold if we are considering the probability space $(\mathbb{R},\mathcal{B},\mu)$, because in that space any probability measure $\mu$ on the Borel $\sigma$-algebra $\mathcal{B}$ is a probability distribution of the random variable $Y:\mathbb{R}\to \mathbb{R}$ defined by $Y(x)=x$. Additionally, a family of probability distributions is a special case of a statistical manifold with the Fisher information matrix as the Riemannian metric, with an affine connection called the $\alpha$-connection (so the definition of a statistical manifold I gave above is a more general definition than Wikipedia's). But in this case $\mathbb{R}$ is orientable so that does not help.
Orientability is merely a convenience, because then positive densities can be identified with volume forms. For a discussion on the non-orientable case in one of the major models of infinite-dimensional statistical manifolds, see the first pages of
Bauer, Martin, Martins Bruveris, and Peter W. Michor. "Uniqueness of the Fisher–Rao metric on the space of smooth densities." Bulletin of the London Mathematical Society 48.3 (2016): 499-506.