Here's an example that shows that $\mathcal{M}_{\geq 0}(M)$ is not closed under addition in general. There may be easier examples, but this is the only one I know.
Let $\pi : \widetilde{M} \to M$ be a covering map. Suppose $g$ is a Riemannian metric on $\widetilde{M}$ such that $f^*g = g$ for all deck transformations $f$. Then there is a Riemannian metric $h$ on $M$ such that $\pi^*h = g$. Note that $s_g = s_{\pi^*h} = \pi^*s_h =s_h\circ \pi$. As $\pi$ is surjective, the functions $s_g$ and $s_h$ are determined by one another.
Now, let $X$ be a non-spin complex surface arising as a complete intersection in some complex projective space (e.g. a smooth hypersurface of $\mathbb{CP}^3$ with odd degree $d \geq 5$), and let $N = S^2\times S^2/\mathbb{Z}_2$ where the $\mathbb{Z}_2$ action is generated by $\sigma(x, y) = (-x, -y)$. Set $M = X\# N$. Then the universal cover of $M$ is $\widetilde{M} = X\# X \#(S^2\times S^2)$, and $\pi : \widetilde{M} \to M$ is a double covering.
In Scalar curvature, covering spaces, and Seiberg-Witten theory, LeBrun showed that $Y(M) < 0$ and $Y(\widetilde{M}) > 0$ where $Y$ denotes the Yamabe invariant. In particular, $\widetilde{M}$ admits a positive scalar curvature metric, while $M$ does not admit a metric with non-negative scalar curvature.
Let $f : \widetilde{M} \to \widetilde{M}$ be the non-trivial deck transformation of $\pi$. If $g$ is a positive scalar curvature metric on $\widetilde{M}$, then $f^*g \neq g$, otherwise there would be a positive scalar curvature metric $h$ on $M$ with $\pi^*h = g$. Note that $f^*g$ is another Riemannian metric on $\widetilde{M}$, and as $s_{f^*g} = s_g\circ f$, it also has positive scalar curvature. Now consider the metric $\tilde{g} := g + f^*g$. As $f^*\tilde{g} = \tilde{g}$, there is a Riemannian metric $h$ on $M$ with $\pi^*h = \tilde{g}$. As $M$ does not admit metrics with non-negative scalar curvature and $s_{\tilde{g}} = s_h\circ\pi$, the metric $\tilde{g}$ does not have non-negative scalar curvature.
So $g, f^*g \in \mathcal{M}_{> 0}(\widetilde{M}) \subset \mathcal{M}_{\geq 0}(\widetilde{M})$, but $g + f^*g \not\in \mathcal{M}_{\geq 0}(\widetilde{M})$.
If you only wanted an example to show (the weaker result) that $\mathcal{M}_{> 0}(M)$ is not closed under addition in general, then this follows from earlier work by Bérard Bergery. He pointed out that there are examples of finite coverings $\pi : \widetilde{M} \to M$ such that $\widetilde{M}$ admits positive scalar metrics, but $M$ doesn't.
For example, one could take $M = (S^2\times\mathbb{RP}^7)\#\Sigma$ where $\Sigma$ is an exotic $9$-sphere with $\alpha(\Sigma) \neq 0$; here $\alpha$ denotes the Hitchin-Lichnerowicz obstruction $\alpha : \Omega^{\text{spin}}_n \to KO_n(\text{pt})$. Hitchin showed that if $M$ is a spin manifold which admits metrics of positive scalar curvature, then $\alpha(M) = 0$; in particular, $\alpha(S^2\times\mathbb{RP}^7) = 0$. Now $M$ is spin and $\alpha(M) = \alpha(S^2\times\mathbb{RP}^7) + \alpha(\Sigma) = \alpha(\Sigma) \neq 0$, so $M$ does not admit metrics of positive scalar curvature. On the other hand $\widetilde{M} = (S^2\times S^7)\#\Sigma\#\Sigma$ is diffeomorphic to $S^2\times S^7$ (because $\Theta_9 = \mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$) which does admit metrics of positive scalar curvature.
In the same way as we did above, we can use this example to construct two metrics $g, f^*g \in \mathcal{M}_{> 0}(S^2\times S^7)$ with $g + f^*g \not\in \mathcal{M}_{> 0}(S^2\times S^7)$.