(This is OP, and I wrote the question with my inactive account. Please excuse.)
I think I found the proof, but there are some points that I am not 100% sure. Yet I think those problems are minor.
Let $X$ be the total space of the locally free sheaf $\mathcal{O}_{\mathbb{P}^n}(-d_1)\oplus \cdots \oplus \mathcal{O}_{\mathbb{P}^n}(-d_r)$.
See this complete Kähler metric I defined here. In short, I defined a codimension $1$ subspace $Y$ inside $U:=(\mathbb{C}^{n+1}\setminus\{0\})\times \mathbb{C}^r$ defined by the equation
$$
F(x,p)= |x_1|^2+\cdots+|x_n|^2-d_1|p_1|^2-\cdots-d_r|p_r|^2=1,
$$
and defined the Riemannian submersion $Y\to X$ under the $U(1)$-action defined by
$$
e^{i\theta} \cdot (x_1,\ldots,x_n,p_1,\ldots,p_r) = (e^{i\theta}x_1,\ldots,e^{i\theta}x_n,e^{-id_1\theta}p_1,\ldots,e^{-i\theta d_r}p_r).
$$
My goal is to show that this metric is bounded geometry.
To proceed, see the answer in this post. It says:
Theorem 1. Let $(M,g)$ be a complete, connected Riemannian manifold. If there is a compact set $K$ in $M$ such that the sectional curvature is positive and bounded above in $M\setminus K$, then $(M,g)$ is bounded geometry.
The proof is beyond my knowledge, so I will skip it.
Our next goal is to compute the sectional curvature of $(X,g)$. For the Riemannian submersion $M\to B$, letting $A$ be the integrability tensor on $TM$,
$$
A_V W = v\nabla^M_{hV} hW + h \nabla^M_{hV} vW,
$$
we have a formula
$$
K^B(V,W)\circ \pi = K^M(V^h,W^h)+ \frac{3g^M(A_{V^h}W^h,A_{V^h}W^h)}{g^B(V\wedge W,V\wedge W)\circ \pi}.
$$
Instead of the Riemannian submersion, with $i:Y\to U$ being the immersion, we may use the submersion from the pullback bundle $i^*TU\to TX$. We still have the same formula (which I am not 100% sure), and $i^*TU$ is flat, we should have
$$
K^B(V,W)\circ \pi = \frac{3g^U(A_{V^h}W^h,A_{V^h}W^h)}{g^B(V\wedge W,V\wedge W)\circ \pi}.
$$
We may assume that $V$ and $W$ are orthonormal. Then,
$$
K^B(V,W)\circ \pi = 3\|A_{V^h}W^h\|.
$$
We let $D:i^*TU\to i^*TU$ be the (1,1)-tensor defined as
$$
DV = \nabla^{U}_V (\mathbf{grad} F).
$$
Then it satisfies the following properties.
Lemma 2.
(1) $D(\partial_{x_j}) = \partial x_j$, $D(\partial_{\overline{x}_j}) = \partial_{\overline{x}_j}$, $D(\partial_{p_j})=-d_j \partial_{p_j}$, $D(\partial_{\overline{p}_j})= -d_j \partial_{\overline{p}_j}.$
(2) $JD = DJ$.
(3) $D$ is flat, that is, $\nabla^{U}_V DW = D\nabla^{U}_V W$.
(4) $\|DV\|\leq \max_j|d_j|\|V\|$.
They are all derived from the direct computation.
We want to show that when $\|\mathbf{grad} F\|\to \infty$, $A\to 0$.
By definition of the integrability tensor, we have
$$
A_{V^h} W^h = v\nabla^{U}_{V^h} W^h = \frac{g^U(\nabla^{U}_{V^h} W^h,\nabla F )}{\|\nabla F\|^2}\nabla F + \frac{g^U(\nabla^{U}_{V^h} W^h, J\nabla F)}{\|\nabla F\|^2}J\nabla F.
$$
By the definition of $\nabla^{U}$, we have
$$
\frac{g^U(\nabla^{U}_{V^h} W^h,\mathbf{grad}F )}{\|\mathbf{grad} F\|^2}\mathbf{grad} F = - \frac{g^U(W^h,\nabla^{U}_{V^h}\mathbf{grad}F \rangle}{\|\mathbf{grad} F\|^2}\mathbf{grad} F = -\frac{g^U( W^h, D(V^h))}{\|\mathbf{grad} F\|^2}\mathbf{grad} F
$$
By Cauchy-Schwartz inequality and (4) of Lemma 2, we have
$$
\left\|\frac{g^U(W^h, D(V^h))}{\|\mathbf{grad}F\|^2}\mathbf{grad} F\right\|\leq \frac{\|D(V^h)\|}{\|\mathbf{grad} F\|}\leq\frac{\sqrt{\max_j |d_j|}}{\|\mathbf{grad} F\|}.
$$
Thus,
$$
\| A_{V^h} W^h \| \leq 2\frac{\sqrt{\max_j |d_j|}}{\|\mathbf{grad}F\|}.
$$
$K^{X}$ is positive. As $\|(\mathbf{x},\mathbf{p})\|\to \infty$, $\|\mathbf{grad}F\|\to \infty$, and the sectional curvature $K^{X}$ goes to zero. Thus, there is a positive real number $R>0$ such that $K^{X}<1$ for all $(\mathbf{x},\mathbf{p})\in Y$ with $\|(\mathbf{x},\mathbf{p})\|>R$. Since $Y \cap \{(\mathbf{x},\mathbf{p})\in U|\|(\mathbf{x},\mathbf{p})\|\leq R\}$ is a compact set, we get the conclusion using Theorem 1.
