$\DeclareMathOperator\Pic{Pic}\DeclareMathOperator\GL{GL}\DeclareMathOperator\GO{GO}$ Let $G$ be a connected reductive group over a field $K$ of characteristic 0. Let $F\subseteq G$ be a connected algebraic subgroup, and set $Y=G/F$. By Proposition 6.10 in Sansuc's paper Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres there is a natural exact sequence of abelian groups $$ {\sf X}^*(G)\to {\sf X}^*(F)\overset{\lambda}{\longrightarrow} \Pic Y\to \Pic G, $$ where ${\sf X}^*(G)$ denotes the character group of $G$, and the map ${\sf X}^*(G)\to {\sf X}^*(F)$ is the restriction homomorphism. In our case $G=\operatorname{GL}_n$, and hence $\Pic G=0$. We obtain a canonical isomorphism $$ \Pic Y\cong \operatorname{coker}\left[{\sf X}^*(G)\to {\sf X}^*(F)\right]. $$ We have a similar isomorphism in the case when $F$ is not necessarily connected. The map $\lambda$ takes a character $\chi\in{\sf X}^*(F)$ to the class of the ${\Bbb G}_m$-torsor $Y'\to Y$, where $Y'=(G\times {\Bbb G}_m)/\chi_*(F)$ and $\chi_*(f)=(f,\chi(f))$ for $f\in F$.
In our case $$ F=\left\{ \begin{pmatrix} A&B\\0&D \end{pmatrix} \ \ \Big |\ \ A\in\GO_k,\ B\in{\rm Mat}_{k,\,n-k}, D\in \GL_{n-k} \right\}. $$ Clearly, $${\sf X}^*(F)={\sf X}^*(\GO_k)\oplus {\sf X}^*(\GL_{n-k}).$$ If $k<n$, we have ${\sf X}^*(\GL_{n-k})\cong {\Bbb Z}$ with generator $\det_{n-k}$. (If $k=n$, then of course ${\sf X}^*(\GL_{n-k})=0$.)
We write $X:={\sf X}^*(\GO_k)$. For $k=1$ we have $X\simeq {\Bbb Z}$. For $2\le k\le n$, the group $X$ is generated by $d=\det_k$ and $c$ with one relation $2d-kc=0$. In other words, $$X\cong{\Bbb Z}^2/\langle (2,-k)\rangle.$$
If $k$ is odd, $k=2p+1$, then the element $(2, -k)\in{\Bbb Z}^2$ is primitive (indivisible), and hence the group $X$ is cyclic. Namely, we consider the following basis of ${\Bbb Z}^2$: $$e_1=2d-(2p+1)c,\quad e_2=d-pc; $$ then $X\cong {\Bbb Z}^2/\langle e_1\rangle \simeq {\Bbb Z}$ with a generator of infinite order $[e_2]$.
If $k$ is even, $k=2p$, then the element $(2, -k)=2(1, -p)\in{\Bbb Z}^2$ is divisible by 2, and hence the group $X$ is isomorphic to ${\Bbb Z}/2{\Bbb Z}\oplus {\Bbb Z}$. Namely, we consider the following basis of ${\Bbb Z}^2$: $$e_1=d-pc,\quad e_2=c;$$ then $X\cong{\Bbb Z}^2/\langle 2e_1\rangle\simeq {\Bbb Z}/2{\Bbb Z}\oplus{\Bbb Z}$ with a generator $[e_1]$ of order 2 and a generator $[e_2]=[c]$ of infinite order.
We assume that $n\ge 2$. If $k<n$, the map ${\sf X}^*(G)\to {\sf X}^*(\GL_{n-k})$ is bijective, and hence $$\operatorname{coker}\left[{\sf X}^*(G)\to {\sf X}^*(F)\right]\simeq {\sf X}^*(\GO_k)=X.$$ Thus for $1\le k<n$ $$ \Pic Y\simeq \begin{cases} {\Bbb Z} &\text{if $k$ is odd;}\\ {\Bbb Z}/2{\Bbb Z}\oplus {\Bbb Z} &\text{if $k$ is even.} \end{cases} $$
For $k=n$ we have ${\sf X}^*(F)=X$ and $$\Pic Y\cong \operatorname{coker}\left[{\sf X}^*(G)\to X\right].$$ Now $$ \operatorname{coker}\left[{\sf X}^*(G)\to X\right]={\Bbb Z}^2/\langle (1,0),(2,-n)\rangle\simeq {\Bbb Z}/n{\Bbb Z} $$ with the generator $[c]$ of order $n$.
Our answers for $k=1$ and $k=n$ coincide with those of OP.