$\newcommand{\RP}{\mathbb{RP}}\newcommand{\ang}[1]{\langle #1\rangle}\newcommand{\R}{\mathbb R}$A representative of
a class in $\Omega_3^O(BO_3)$ is a 3-manifold $M$ together with a principal $O_3$-bundle $P\to M$. Principal
$O_3$-bundles are equivalent to rank-3 real vector bundles, so I will use 3-manifolds $M$ together with rank-3
real vector bundles $E\to M$. In this setting, your invariants correspond to products of Stiefel-Whitney numbers of
$M$ and $E$, evaluated on the mod 2 fundamental class of $M$.

Specifically, I will use the more standard notation $g^3 = \ang{w_1(E)^3, [M]}$, $gw_2'(V_{SO(3)}) =
\ang{w_1(E)w_2(E), [M]}$, $w_3'(V_{SO(3)}) = \ang{w_3(E), [M]}$, and $gw_1(T)^2 = \ang{w_1(E)w_1(M)^2, [M]}$. For
brevity I'll write $w_1(E)^3$ for $\ang{w_1(E)^3, [M]}$, etc., and let $S(M,E)$ denote the tuple $(w_1(E)^3,
w_1(E)w_2(E), w_3(E), w_1(E)w_1(M)^2)$.

Let $\ell_{\RP^n}\to\RP^n$ denote the tautological line bundle (I'll also do this for $\RP^1 = S^1$); if $x\in H^1(\RP^n;\mathbb Z/2)$ denotes the
generator, then $w(\ell_{\RP^n}) = 1+x$. Then, a generating set for
$\Omega_3^O(BO_3)$ is

$$\{ (\RP^3, \ell_{\RP^3}^{\oplus 3}), (\RP^3, \ell_{\RP^3} \oplus\underline\R^2), (S^1\times\RP^2,
\ell_{S^1}\oplus\underline\R^2), (S^1\times\RP^2, \ell_{S^1}\oplus\ell_{\RP^2}\oplus\underline\R)\}. $$

Specifically, using the Whitney sum formula one can calculate that

- $S(\RP^3, \ell_{\RP^3}^{\oplus 3}) = (1, 1, 1, 0)$,
- $S(\RP^3, \ell_{\RP^3}\oplus\underline\R^2) = (1, 0, 0, 0)$,
- $S(S^1\times\RP^2, \ell_{S^1}\oplus\underline\R^2) = (0, 0, 0, 1)$, and
- $S(S^1\times\RP^2, \ell_{S^1}\oplus\ell_{\RP^2}\oplus\underline\R) = (1, 1, 0, 1)$.

These four vectors are linearly independent in $\mathbb F_2^4$, so these manifolds generate the bordism group.

If you want manifolds equal to $1$ on one invariant and $0$ on the others, you can take disjoint unions of these
manifolds: for example, $(0, 1, 0, 0)$ is represented by the disjoint union of the last three generators. It may be
possible to find simpler representatives.