I don't know how to answer this question completely, but I'll make some observations.

The oriented vector bundles over $M$ are classified by homotopy classes of maps $f:M\to \tilde{G}_5$, the oriented Grassmannian, denoted $\pi_0 ([M, \tilde{G}_5])$, where $[X,Y]$ denotes the space of mappings from $X$ to $Y$. Let $E\to M$ be a 5-dimensional vector bundle over $M$. Then the Euler class $e(E)=0$, and hence there is a global section. So $E=F\oplus \varepsilon_1$, where $\varepsilon_1=M\times \mathbb{R}$ is the trivial bundle, and $F\to M$ is a 4-dimensional oriented vector bundle. Hence $\pi_0 [M,\tilde{G}_4]\to \pi_0 [M,\tilde{G}_5]$ is onto. I'm not sure if this helps, but it is related to the observation made in the paper that you linked to that in the case of spin 5-manifolds, this bundle is quaternionic.

A second observation is that $M$ admits a cell structure with one $5$-cell. The 4-skeleton $M^4\subset M$ is a cofibration (note that up to homotopy, $M^4 \simeq M - pt.$, so is well defined up to homotopy). Hence the map $[M, \tilde{G}_5]\to [M^4,\tilde{G}_5]$ is a fibration, with fiber $[(S^5,\ast),(\tilde{G}_5,\ast)]$. The fiber has two components, in bijection with the components of $[S^5,\tilde{G}_5]$ from the homotopy exact sequence and the fact that $\tilde{G}_5$ is simply-connected for the fibration $[(S^5,\ast),(\tilde{G}_5,\ast)]\to [S^5,\tilde{G}_5]\to \tilde{G}_5$.

A third observation is that by cellular approximation, $\pi_0[M^4,\tilde{G}_5] = \pi_0[M^4, \tilde{G}_5^5]$, where $\tilde{G}_5^5$ is the 5-skeleton. This is because any map $f:M^4 \to \tilde{G}_5$ may be homotoped to $f: M^4\to \tilde{G}_5^4$. Moreover, homotopies between such maps may be homotoped to $\tilde{G}_5^5$, since $M\times [0,1]$ is 5-dimensional.

A fourth observation is that $\tilde{G}_5^5 \cong \tilde{G}_\infty^5$, where $\tilde{G}_\infty$ is the infinite oriented Grassmannian obtained as the nested union $\tilde{G}_1\subset \tilde{G}_2 \subset \cdots$, where each inclusion is induced by stabilization. This can be proved using the cell structure on Grassmannians given by Schubert cells, but I suspect there may be a more direct approach.
Then $\pi_0 [M^4,\tilde{G}_5]=\pi_0[M^4,\tilde{G}_5^5]=\pi_0[M^4,\tilde{G}_\infty^5] = \pi_0[M^4,\tilde{G}_\infty] =\pi_0 [M^4,BSO]\leq \tilde{KO}(M^4)$, the real reduced K-theory of $M^4$. The image of $\pi_0 [M^4,\tilde{G}_\infty] \to \pi_0[M^4,G_\infty]=\tilde{KO}(M^4)$ should correspond to the components where $im \{\pi_1(M^4)\to \pi_1(G_\infty)\}=0$, where $\pi_1(G_\infty)=\mathbb{Z}/2$. In principle, this should be computable from the Atiyah-Hirzebruch spectral sequence and the cohomology of $M^4$, which is determined by the cohomology of $M$.

A fifth observation is that with the fibration $[(S^5,\ast),(\tilde{G}_5,\ast)]\to [M, \tilde{G}_5] \to [M^4,\tilde{G}_5]$,
the components in the image will correspond to maps where $im\{\pi_0 [M^4,\tilde{G}_5]\to \pi_0 [S^4,\tilde{G}_5]\} =0$, where $S^5\subset M^4$ is the attaching map of $\partial B^5$, the 5-handle. Again, this should be determined by the map $\tilde{KO}(M^4) \to \tilde{KO}(S^4)$.

Now for the above fibration, the fiber has two components, so the number of components should be at most twice the number of components of the base (determined by the above K-theory computations). I'm not sure how to determine the image $\pi_1 [M^4,\tilde{G}_5]\to \pi_0 [(S^5,\ast),(\tilde{G}_5,\ast)]$.

The upshot is that for each 5-dimensional vector bundle over $M$, one may modify it by "inserting" a copy of the tangent bundle to $S^5$ (this can be descibed in terms of clutching functions of the attaching map of the 5-cell). Then the above question reduces to whether this changes the vector bundle up to isomorphism or keeps it the same? I think the above description classifies them up to this operation.

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