= Solution
Here is an intrinsic definition that also proves well-definedness. For a <complex line bundle> $L$, define its <First Chern class> to be the <Euler class> of its canonically oriented underlying real rank-two bundle: pull its integral <Thom class> back along the zero section after forgetting relative support. The complex orientation fixes the sign, so this construction makes no arbitrary choice of generator.
For a rank-$r>0$ <complex vector bundle> $E$, let $\pi:\mathbb P(E)\to X$ be its <projective bundle> of lines, let $S\subset\pi^*E$ be the <tautological bundle>, and put $h=-c_1(S)=c_1(S^*)$. On every fiber $\mathbb{CP}^{r-1}$, $h$ is the positive degree-two generator. The <Leray-Hirsch theorem> says that if globally defined <cohomology> classes restrict to a free basis of the <cohomology> of every fiber, then multiplication by those classes identifies the <cohomology> of the total space with a <free module> over the base. Applied here, it gives
$$
\bigoplus_{j=0}^{r-1}H^{*-2j}(X;\mathbb Z)\xrightarrow{\ \cong\ }H^*(\mathbb P(E);\mathbb Z),\qquad(a_j)_j\longmapsto\sum_j\pi^*a_j\smile h^j.
$$
The finite trivializing cover in the question is sufficient for this application: the assertion holds on each trivializing open set by the <Künneth theorem>, since the fiber has finite free <cohomology>, and the <Mayer–Vietoris sequence> and the <Five lemma> glue it over the finite cover. The same local argument constructs the oriented <Thom class> used for line bundles. It does not require a choice of classifying map.
There are therefore unique classes $c_i(E)\in H^{2i}(X;\mathbb Z)$ such that
$$
\boxed{h^r+\pi^*c_1(E)h^{r-1}+\cdots+\pi^*c_r(E)=0.}
$$
Define $c_0(E)=1$ and $c_i(E)=0$ for $i>r$. For rank zero the <Total Chern class> is $1$. This is the <projective bundle definition of Chern classes>. Existence and uniqueness follow by expressing $h^r$ in the displayed <free module> basis, with degrees determining each coefficient. The <projective bundle> and <tautological bundle> are intrinsic to $E$, and the line <Thom class> is uniquely fixed by orientation. Hence the resulting <Chern classes> do not depend on trivializations or other auxiliary choices. For a line bundle the relation is $h+c_1(E)=0$, agreeing with the original normalization. Pulling back this unique relation also proves naturality. This standard construction and the sum theorem are treated in https://pi.math.cornell.edu/~hatcher/VBKT/VB.pdf[Vector Bundles and K-Theory, Section 3.1].
The requested result is the <Whitney sum formula for Chern classes>:
$$
\boxed{c(E\oplus E')=c(E)c(E'),\qquad c_k(E\oplus E')=\sum_{i+j=k}c_i(E)\smile c_j(E').}
$$
Here $c(E)=1+c_1(E)+\cdots+c_r(E)$, and the formula holds for complex vector bundles over a common base. In particular, a trivial bundle has total Chern class $1$.
Now take $B=\mathbb{CP}^{n-1}$ and let $L\subset B\times\mathbb C^n$ be its <tautological bundle>. The standard <Hermitian inner product> gives the rank-$(n-1)$ complex bundle $Q=L^\perp$, with $L\oplus Q\cong\underline{\mathbb C}^{\,n}$. The <orthogonal complex line flag manifold> in the question is precisely $\mathbb P(Q)$: over a first line $\ell$, the second line is any line in $\ell^\perp$. Local orthonormal frames give this identification as a <fiber bundle>, with fiber $\mathbb{CP}^{n-2}$.
Let $x=c_1(L^*)$ on $B$, also writing $x$ for its pullback to $\mathbb P(Q)$. By part 3(a), $H^*(B;\mathbb Z)=\mathbb Z[x]/(x^n)$. The <Whitney sum formula for Chern classes> gives
$$
(1-x)c(Q)=1,\qquad c(Q)=1+x+x^2+\cdots+x^{n-1},\qquad c_i(Q)=x^i.
$$
Let $S_2$ be the second tautological line on $\mathbb P(Q)$, and set $y=c_1(S_2^*)$. Thus $x,y$ are exactly the pullbacks of the positive hyperplane classes from the two projective factors. The <projective bundle definition of Chern classes> for $Q$ gives
$$
y^{n-1}+xy^{n-2}+x^2y^{n-3}+\cdots+x^{n-1}=0.
$$
Together with $x^n=0$, this gives a surjective graded ring map
$$
\mathbb Z[x,y]\big/(x^n,\,x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})\longrightarrow H^*(X;\mathbb Z).
$$
There are \b[no additional relations]. Indeed the second relation is monic of degree $n-1$ in $y$, so <polynomial division> makes its source free over $\mathbb Z[x]/(x^n)$ on $1,y,\ldots,y^{n-2}$. The <projective bundle formula for complex vector bundles> gives exactly the same free basis on the target. The map sends each basis element to its corresponding basis element, and is therefore an isomorphism. Consequently
$$
\boxed{H^*(X;\mathbb Z)=\mathbb Z[x,y]/\left(x^n,\ \sum_{j=0}^{n-1}x^{n-1-j}y^j\right),\qquad |x|=|y|=2.}
$$
This is the <cohomology ring of the orthogonal complex line flag manifold>. Its additive basis is $x^iy^j$ with $0\leq i<n$, $0\leq j<n-1$. As a symmetry check, multiplying the second relation by $y-x$ gives $y^n-x^n=0$, so $y^n=0$ as expected from the second projection. For $n=2$, every line has a unique orthogonal line, and the relations become $x^2=0$, $x+y=0$, giving the ring of $\mathbb{CP}^1$. If $n=1$, the space is empty and the second relation is $1=0$, so the printed formula still gives the zero <cohomology ring>.
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