The answer to your question, as I now understand it, is no. In particular, we can construct a matrix of you particular pattern with a positive determinant that fails to be positive definite.
In particular, consider the matrix
$$
M =
\pmatrix{
1&-1&-1&0&0&0\\
-1&1&-1&0&0&0\\
-1&-1&1&0&0&0\\
0&0&0&1&-1&-1\\
0&0&0&-1&1&-1\\
0&0&0&-1&-1&1\\
}
$$
which has eigenvalues $-1-1,2,2,2,2$
Claim: The rank of $M$ is less or equal to $2k_1$.
One way to see this is by using the Schur product $(A\odot B)$.
Let $u\in \mathbb{R}^{k_1+k_2}$ be the column vector with all coordinates equal to 1.
Lemma: Let $F\in \mathbb{R}^{k_1+k_2\times k_1+k_2}$ be a psd matrix such that $u\in \operatorname{Image}(F)$. If $G\in \mathbb{R}^{k_1+k_2\times k_1+k_2}$ is another psd matrix then $\operatorname{rank}(G\odot F)\geq \operatorname{rank}(G)$.
Proof of the Lemma:
There is $\epsilon>0$ such that $F-\epsilon uu^t$ is psd.
Since $G$ is also psd, $G\odot (F-\epsilon uu^t)$ is psd by a well known property of the Schur product.
Now, $$G\odot F=G\odot (F-\epsilon uu^t)+\epsilon\ G\odot uu^t=G\odot (F-\epsilon uu^t)+\epsilon\ G.$$
Finally, since $G\odot (F-\epsilon uu^t)$ and $G$ are psd, $\operatorname{rank}(G\odot F)\geq \operatorname{rank}(G)$.$\square$
Proof of the Claim:
Let $E=\begin{pmatrix} 1_{k_1\times k_1} & 0 \\ 0 & 1_{k_2\times k_2} \end{pmatrix}$, where $1_{k_i\times k_i}$ is the matrix of order $k_i$ with all coordinates equal to 1.
Note that $u\in \operatorname{Image}(E)$, $E$ is psd and $M\odot E=\begin{pmatrix} A & 0 \\ 0 & C\end{pmatrix}$.
Then, by our Lemma, $2k_1=\operatorname{rank}(M\odot E)\geq \operatorname{rank}(M)$. $\square$
Best Answer
At least a lower-bound is possible; in fact, the following holds for any PSD matrix $S$: \begin{align} (1+\|S\|_{\mathrm{op}})^r \geq \det(I_n +S), \end{align} where $\|S\|_{\mathrm{op}} \triangleq \sup_{x\in \mathbb{R}^n\setminus\{0\}} \|S x\|_2/\|x\|_2$ is the operator norm of $S$.
Proof. Since $S$ is PSD, there exists an orthogonal matrix $U\in \mathbb{R}^{n\times n}$, and $\lambda_1\geq \dots \geq \lambda_n\geq 0$ such that $$S= U\mathrm{diag}(\lambda_1,\dots,\lambda_n) U^\top.$$ In this case, the eigenvalues of $S$ are $\lambda_1,\dots, \lambda_n$ in decreasing order. This implies that \begin{align} \det(I_n +S)&= \det(U (I_n+\mathrm{diag}(\lambda_1,\dots, \lambda_n) )U^\top), \\ &= \det(U) \det(I_n +\mathrm{diag}(\lambda_1,\dots, \lambda_n)) \det(U^\top),\\ &=\det(I_n +\mathrm{diag}(\lambda_1,\dots, \lambda_n)),\\ & =\prod_{i=1}^n(1+\lambda_i),\hspace{6cm} (1) \end{align} where the penultimate equality follows by the fact that $\det(U)\det(U^\top)=1$, since $U$ is orthogonal. Now, since $S$ has rank $r$, we must have $\lambda_{n-r+1}=\dots=\lambda_{n}=0$. Furthermore, by the definition of the operator norm, we have $\|S\|_{\mathrm{op}}=\lambda_1\geq \dots\geq \lambda_n$. This, together with (1) implies that \begin{align} \det(I_n +S) &=\prod_{i=1}^r(1+\lambda_i),\\ & \leq (1+\|S\|_{\mathrm{op}})^r. \end{align}