How few $k$-dimensional subspaces of $V$ are enough to have a complement to each $n-k$-dimensional subspace?

If $F$ is algebraically closed then $d(n,k)=k(n-k)+1$.

For $W\subset V$ of dimension $k$, write $X_W\subset Gr(V,n-k)$ for the set of $n-k$-dimensional subspaces of $V$ that intersect $W$ non-trivially. Then $X_W$ is a subvariety of codimension $1$, and $\{W_i\}$ is a complement repository iff $\cap X_{W_i}=\varnothing$.

Claim: If $A\subset Gr(V,n-k)$ is a non-empty Zariski closed subset, then there exists $W$ such that $\dim(A\cap X_{W})\leq\dim(A)-1$.

Proof: Suppose $A$ is irreducible. Then $\dim(A\cap X_{W})\leq \dim(A)-1$ if and only if $A\not\subset X_W$. Define $$ Y_A:=\left\{W\in Gr(V,k): A\not\subset X_W\right\}. $$ Certainly $Y_A$ is non-empty (it contains any complement of any point in $A$), and $Y_A$ is Zariski open. So in the case $A$ is irreducible, we may take $W$ to be any element of $Y_A$.

If $A$ is not irreducible, there is a decomposition into irreducible components $A=\cup A_i$. Then $\cap Y_{A_i}$ is non-empty (it is a finite intersection of non-empty Zariski open sets), and every $W\in \cap Y_{A_i}$ has the desired property.$\blacksquare$

By induction, we can find a sequence $W_1$, $W_2$, $\ldots$, $W_{k(n-k)+1}$ such that $$ \dim\left(\bigcap_{i} X_{W_i}\right)\leq \dim(Gr(V,n-k))-\big(k(n-k)+1\big)=-1. $$ Thus $\cap X_{W_i}=\varnothing$, so $\{W_i\}$ is a complement repository of size $k(n-k)+1$.

Here is a particular case showing that $d(n,k)=d_K(n,k)$ definitely depends on the field $K$ in general, beyond pathologies of finite fields:

$d_{\mathbf{R}}(4,2)=4$

(although $d(4,2)=5$ in the algebraically closed case).

Possibly a similar argument yields $d_K(4,2)=4$ as soon as $K$ has non-squares but I haven't checked details.

Consider the 2-planes in $K^4$ $$A=\{(t,s,0,0)\},\;B=\{(0,0,t,s)\},\; C=\{(t,s,t,s)\},\;D=\{(t,t+2s,t+s,s)\},$$ where $(t,s)$ is understood to range over $K^2$. To prove $d_{\mathbf{R}}(4,2)\le 4$ ($d_K(4,2)\ge 4$ is an elementary verification for an arbitrary field $K$), let's prove that

$\{A,B,C,D\}$ is a complement repository as soon as $-1$ is not a square in $K$.

Proof:

For the moment, $K$ is an arbitrary field.

For $(e,f)\in K^2\smallsetminus \{(0,0)\}$, define $P_{e,f}$ to be the 2-plane $\{(te,tf,se,sf):(t,s)\in K^2\}$. Write $\mathcal{P}=\{P_{e,f}:(e,f)\neq (0,0)\}$.

Clearly $P_{e,f}$ intersects nontrivially each of $A,B,C$ (taking $s=0$, $t=0$, $t=s$ respectively). I claim that conversely if a 2-plane $P$ intersects nontrivially each of $A,B,C$, then $P\in\mathcal{P}$. This is a simple exercise. Indeed, if $P$ intersects $A,B$ nontrivially, then $P=(P\cap A)\oplus (P\cap B)$ and hence there exists $(c,f,a,b)$ with $(c,f),(a,b)\neq (0,0)$ such that $P=\{(tc,tf,sa,sb):(t,s)\in K^2\}$. If $P\cap C$ is nontrivial, for some $(t,s)\neq (0,0)$ we have $tc=sa$ and $tf=sb$. This means that both $(a,c)$ and $(b,f)$ are collinear to the nonzero vector $(t,s)$. If $(b,f)=(0,0)$ we deduce $P=P_{1,0}$. Otherwise, $(a,c)=e(b,f)$ for some $e\in K$. So we have $P=\{(tef,tf,seb,sb):(t,s)\in K^2\}=P_{e,1}$.

Therefore, the final claim that $\{A,B,C,D\}$ is a complement repository is equivalent to showing that $D\cap P_{e,f}=\{0\}$ for every $(e,f)\in K^2 \smallsetminus\{(0,0)\}$.

Since a basis of $P_{e,f}$ is $((e,f,0,0),(0,0,e,f))$ and a basis of $D$ is $((0,2,1,1),(1,1,1,0))$, this amounts to showing that the determinant $\delta(e,f)=\begin{vmatrix}e & f & 0 & 0\\0 & 0 & e & f\\ 0 & 2 & 1 & 1\\ 1 & 1 & 1 & 0\end{vmatrix}$ is nonzero for all $(e,f)\in K^2\smallsetminus\{(0,0)\}$. The computation yields $\delta(e,f)=e^2+f^2$.

Therefore the non-vanishing holds if and only if $-1$ is not a square in $K$. This means that $\{A,B,C,D\}$ is a complement repository if and only if $-1$ is not a square in $K$.

See these papers:

Covering by Complements of Subspaces, II., W. Edwin Clark and Boris Shekhtman, Proc. Amer. Math. Soc.125 (1997), no. 1, 251--254. (link here, unrestriced access; MR review)

Covering by Complements of Subspaces, W. Edwin Clark and Boris Shekhtman, Linear and Multilin. Algebra, Vol 49,1995, pp. 1--13. (link here, restricted accessMR review)