= Solution
Write $u=u_n\in H^n(K(\mathbb Z/2,n);\mathbb F_2)$ for the fundamental class. We use the path-loop <Serre spectral sequence>, its multiplicative structure, and the <Kudo transgression theorem>: transgression of the fundamental classes and their compatible <Steenrod squares> is natural, equivalently <cohomology suspension> commutes with these stable operations. The initial ring is $H^*(K(\mathbb Z/2,1);\mathbb F_2)=\mathbb F_2[t_1]$.
Here is the low-degree calculation. In the first path fibration, $t$, $t^2$ and $t^4$ transgress respectively to $u_2$, $\operatorname{Sq}^1u_2$ and $\operatorname{Sq}^2\operatorname{Sq}^1u_2$, in degrees $2,3,5$. These supply all indecomposable base classes through degree five; the other classes there are products. Repeating the path-loop calculation raises the degree of the transgressive classes by one. For $n=3$, the class $\operatorname{Sq}^2u_3$ transgresses from $u_2^2$, and $\operatorname{Sq}^2\operatorname{Sq}^1u_3$ from the degree-five generator. For $n=4$, $\operatorname{Sq}^3u_4$ transgresses from $u_3^2$. For larger $n$ all four displayed low-degree operations are transgressive indecomposables. Acyclicity of the path-space total <cohomology> forces these transgressions and excludes additional classes in this range.
More systematically, this is the range up to $n+3$ of the <Serre polynomial generators for mod-two Eilenberg–MacLane cohomology>:
$$
H^*(K(\mathbb Z/2,n);\mathbb F_2)
=\mathbb F_2[\operatorname{Sq}^Iu_n:
I\text{ admissible},\ e(I)<n].
$$
For an <admissible sequence of Steenrod squares>, $I=(i_1,\ldots,i_r)$ has $i_j\geq2i_{j+1}$; its degree increment is $\sum i_j$, and its <Steenrod excess> is $e(I)=i_1-i_2-\cdots-i_r$. Include the empty sequence. Up to increment three the nonempty possibilities are $(1),(2),(3),(2,1)$; the strict excess bound and possible products explain precisely the small-$n$ exceptions.
\b[The <low-degree mod-two cohomology of an Eilenberg–MacLane space> is]
$$
\boxed{
H^i=
\begin{cases}
\mathbb F_2\{1\},&i=0,\\
0,&0<i<n,\\
\mathbb F_2\{u\},&i=n,\\
\mathbb F_2\{\operatorname{Sq}^1u\},&i=n+1,\\
\mathbb F_2\{\operatorname{Sq}^2u\},&i=n+2,\\
\mathbb F_2\{u\operatorname{Sq}^1u,\,
\operatorname{Sq}^2\operatorname{Sq}^1u\},&i=n+3,\ n=2,\\
\mathbb F_2\{\operatorname{Sq}^3u,\,
\operatorname{Sq}^2\operatorname{Sq}^1u\},&i=n+3,\ n\geq3.
\end{cases}}
$$
Negative-degree <cohomology> is zero. For $n=2$, $\operatorname{Sq}^2u=u^2$ and $\operatorname{Sq}^3u=0$ by instability, so the product in degree five must not be omitted. For $n=3$, $\operatorname{Sq}^3u=u^2$ is a product, still independent from $\operatorname{Sq}^2\operatorname{Sq}^1u$. For $n\geq4$ the listed classes are indecomposable. The <Adem relations> include $\operatorname{Sq}^1\operatorname{Sq}^1=0$ and $\operatorname{Sq}^1\operatorname{Sq}^2=\operatorname{Sq}^3$, so no further increment-three class comes from reversing the two squares.
Now use the space actually printed in the PDF,
$$
Y=\mathbb{RP}^{10}/\mathbb{RP}^{6};
$$
the converted TeX dropped the projective-space $P$'s. This <stunted real projective space> has one cell in dimensions $7,8,9,10$, besides its basepoint, and is $6$-connected. The integral cellular boundary is $2$ in even dimensions and zero in odd dimensions. Hence
$$
\widetilde H_i(Y;\mathbb Z)=
\begin{cases}\mathbb Z/2,&i=7,9,\\0,&\text{otherwise}.\end{cases}
$$
The <Hurewicz theorem> gives $\pi_7(Y)=\mathbb Z/2$.
For the next two groups, let $K=K(\mathbb Z/2,7)$ and choose $f:Y\to K$ representing the generator of $H^7(Y;\mathbb F_2)$. It induces an isomorphism on $\pi_7$, and the target's higher <homotopy groups> vanish.
We need the <low-degree integral homology of a mod-two Eilenberg–MacLane space>. Here it is
$$
H_7(K;\mathbb Z)=\mathbb Z/2,\quad H_8(K;\mathbb Z)=0,\quad
H_9(K;\mathbb Z)=\mathbb Z/2,\quad H_{10}(K;\mathbb Z)=\mathbb Z/2.
$$
To justify the torsion orders, the <Serre class> theorem first makes every positive-degree integral <homology> group of $K$ a finite 2-group. The <universal coefficient theorem for cohomology> with $\mathbb F_2$ and the low-degree dimensions give one cyclic summand in degrees seven, nine and ten and none in degree eight. The mod-two <Bockstein homomorphism> is $\operatorname{Sq}^1$. Its relevant nonzero actions are
$$
u\mapsto\operatorname{Sq}^1u,\qquad
\operatorname{Sq}^2u\mapsto\operatorname{Sq}^3u,\qquad
\operatorname{Sq}^2\operatorname{Sq}^1u
\mapsto\operatorname{Sq}^3\operatorname{Sq}^1u.
$$
The last target is nonzero: $(3,1)$ is admissible with excess two, below seven. It also follows by transgressing the nonzero square $(\operatorname{Sq}^1u_2)^2$ through successive path fibrations. Each nonzero Bockstein pairs the mod-two classes associated with a cyclic integral summand of order exactly two; for a summand of order $2^r$ with $r>1$, this first Bockstein would be zero. In degree ten, $\operatorname{Sq}^3u$ is already the Ext class from $H_9$, so the independent class $\operatorname{Sq}^2\operatorname{Sq}^1u$ detects $H_{10}$. This proves all four integral groups without confusing them with mod-two Betti numbers.
Let $x\in H^1(\mathbb{RP}^{10};\mathbb F_2)$ be its usual generator. The quotient classes in degrees $7$ through $10$ identify with $x^7,\ldots,x^{10}$ via the pair's <cohomology>. The <Steenrod squares on real projective space> satisfy
$$
\operatorname{Sq}^j(x^r)=\binom rj x^{r+j}.
$$
Thus
$$
f^*u=x^7,\quad
f^*\operatorname{Sq}^1u=x^8,\quad
f^*\operatorname{Sq}^2u=x^9,\quad
f^*\operatorname{Sq}^3u=x^{10},\quad
f^*\operatorname{Sq}^2\operatorname{Sq}^1u=0,
$$
since $\binom72$ and $\binom73$ are odd, while $\binom82$ is even. In degree nine, the nonzero <cohomology> map detects the map on $H_9$, as $H_8=0$ on both sides. Consequently $f_*:H_9(Y;\mathbb Z)\to H_9(K;\mathbb Z)$ is an isomorphism.
Treat $f$ as a mapping-cylinder pair $(K,Y)$. It is $8$-connected: both spaces are $6$-connected, $\pi_7$ is an isomorphism, and $\pi_8(K)=0$. The relative <homology> sequence gives $H_9(K,Y)=0$, since $H_9(Y)\to H_9(K)$ is an isomorphism and $H_8(Y)=0$. The <Relative Hurewicz theorem> then gives $\pi_9(K,Y)=0$, and the relative <homotopy> sequence identifies this group with $\pi_8(Y)$. Thus $\pi_8(Y)=0$ and the pair is now $9$-connected.
Next,
$$
H_{10}(K,Y)\cong H_{10}(K)\cong\mathbb Z/2,
$$
because $H_{10}(Y)=0$ and the degree-nine map is an isomorphism. Apply the <Relative Hurewicz theorem> again and use $\pi_{10}(K)=\pi_9(K)=0$:
$$
\pi_9(Y)\cong\pi_{10}(K,Y)\cong H_{10}(K,Y)\cong\mathbb Z/2.
$$
Therefore \b[the <homotopy groups of the stunted projective space through degree nine> are]
$$
\boxed{\pi_i(\mathbb{RP}^{10}/\mathbb{RP}^{6})=
\begin{cases}
0,&1\leq i\leq6,\\
\mathbb Z/2,&i=7,\\
0,&i=8,\\
\mathbb Z/2,&i=9.
\end{cases}}
$$
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