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 by 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 Zariksi 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 $k(n-k)+1$.

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 $k(n-k)+1$.