$\newcommand{\Pin}{\mathrm{Pin}}\newcommand{\Sq}{\mathrm{Sq}}$
To simplify notation, I'll denote $(\Pin_n^\pm\times\mathrm{SU}_2)/(\mathbb Z/2)$ by $\Pin_n^{h\pm}$, since these are
analogues of the groups $\mathrm{Spin}_n^h = (\mathrm{Spin}_n\times\mathrm{SU}_2)/(\mathbb Z/2)$. Analogously to how a
spin^{$c$}-structure on a manifold $M$ determines a line bundle $L$ with $w_2(L) = w_2(M)$, a
pin^{$h\pm$}-structure on $M$ determines a principal $\mathrm{SO}_3$-bundle $Q$ with $w_2(Q) = w_2(M)$ (for
pin^{$h+$}) or $w_2(Q) = w_2(M) + w_1^2(M)$ (for pin^{$h-$}).

This question is actually several questions in one: the argument differs for $\Pin^{h+}$ and $\Pin^{h-}$, and it's also necessary
to specify the maps $\Omega_5^{\Pin^{h\pm}}\to\Omega_5^O(K(\mathbb Z/2, 2))$. Any characteristic class of $M$ or
$Q$ in $H^2(M;\mathbb Z/2)$ determines a natural map to $K(\mathbb Z/2,2)$, and hence a map between these bordism
groups. There are several examples, such as $w_2(Q)$ and $w_2(Q) + w_1^2(M)$, and the answer could in principle differ depending
on which class one chooses.

As such, I have a partial answer: *in the case of pin*^{$h+$} 5-manifolds where the map to $K(\mathbb Z/2,2)$
is $w_2(Q)$ or $w_2(Q) + w_1^2(M)$, the answer is yes, the invariant vanishes.

Since $u_2\mathrm{Sq}^1 u_2 + \mathrm{Sq}^2\mathrm{Sq}^1u_2$ is a cobordism invariant, it suffices to check this on
a generator of $\Omega_5^{\Pin^{h+}}$. This group is isomorphic to $\mathbb Z/2$ (this is due to
Freed-Hopkins, Theorem 9.89), so we need only to find a single nonbounding
pin^{$h+$} 5-manifold and check there.

This generator is the Wu manifold $W := \mathrm{SU}_3/\mathrm{SO}_3$, with the principal $\mathrm{SO}_3$-bundle $Q$
given by the quotient map $\mathrm{SU}_3\twoheadrightarrow W$. It's known that $H^*(W;\mathbb Z/2) =
\mathbb Z/2[z_2,z_3]/(z_2^2, z_3^2)$ with $|z_i| = i$; its nonzero Stiefel-Whitney classes are $w_2(W) = z_2$ and $w_3(W)
= z_3$, and that $w_2(Q) = z_2$ (this is discussed, for example, at the end of section 6 of Xuan Chen's
thesis). In particular, $w_2(W) = w_2(Q)$, so $W$
admits a pin^{$h+$}-structure whose corresponding $\mathrm{SO}_3$-bundle is $Q$, and because $\langle
w_2(Q)w_3(W), [W]\rangle = 1$, it doesn't bound as a pin^{$h+$} manifold. Therefore it generates
$\Omega_5^{\Pin^{h+}}$.

Using the Wu formula, $\Sq^1z_2 = z_3$ and $\Sq^2z_3 = z_2z_3$, so

$$ w_2(Q)\Sq^1w_2(Q) + \Sq^2\Sq^1w_2(Q) = 2z_2z_3 = 0,$$

so we conclude that this invariant vanishes for all pin^{$h+$} 5-manifolds. (Since $w_1(W)^2 = 0$, we can
also conclude this for the map to $K(\mathbb Z/2,2)$ given by $w_2(Q) + w_1(M)^2$.)

An analogous argument should be possible in the pin^{$h-$} case, but since
$\Omega_5^{\Pin^{h-}}\cong\mathbb Z/2\oplus\mathbb Z/2$ (this is again Freed-Hopkins, Theorem 9.89), one would have
to find two linearly independent generators and check on them. $W$ also admits a pin^{$h-$}-structure with
corresponding bundle $Q$, but I wasn't able to figure out what the other generator is.