I have no trouble to those integral except this one
$$ \iint_Rf(x,y)\,dA=V=\int^b_a\int^d_cf(x,y) \,dy\,dx\quad\text{R here indicate a region}$$
I agree with $$A(x)=\int^d_c f(x,y)\,dy=\int^d_c\,dA(x)$$
Now when you try to obtain $V=\int^b_aA(x)\,dx$, isn't this suppose to be something like $$\int^b_a\int^d_c\,dA(x)\,dx$$ I am having trouble to obtain $$\iint_Rf(x,y)\,dA$$when you have $dA$, where is another $f(x,y)$ comes from? isn't $dA$ indicating integration of $f(x,y)$ wrt $dy$ already. It just seems not algebraically consistent. I've seen other source that taking $dA=dxdy$, but that not consistent with my undersanding.
Thank you for any commments
The "$A$" in $A(x) = \int_c^d f(x,y)\,dy$ is a different area than the $A$ in $\iint_R f(x,y)\,dA$.
$R = [a, b]\times[c,d]$ is a rectangle in $\Bbb R^2$, which we can identify with the $xy$-plane in $\Bbb R^3$. Under this identification $R \equiv [a, b]\times[c,d]\times 0$. All the integrals are calculating the volume under the function $f(x,y)$, or a stepping stone to calculate that volume. That is, the volume of the solid $S = \{(x,y,z) \mid (x,y) \in R, 0 \le z \le f(x,y)\}$.
In $\iint_R f(x,y)\,dA$, the $dA$ is a differential of area in $R$. That is (in the Riemann interpretation), $R$ is broken up into many tiny regions, each so small that $f$ is nearly constant on them. $dA$ is the area of one these tiny regions in the $xy$-plane. On each of these regions, choose a point $(x,y,0)$ in the region and $f(x,y)$ is the height of the column over the region below $f$. Then $f(x,y)dA$ is an estimation of the volume of $S$ above the tiny region. Sum up the volumes of the columns over all the tiny regions to estimate the volume of $S$, which converges to that volume as the regions become smaller. So the $A$ here is area in the $xy$-plane.
On the other hand, in $A(x) = \int_c^d f(x,y)\,dy, x$ is held constant, so let me replace it with $x_0: A(x_0) = \int_c^d f(x_0,y)\,dy$ This frees up $x$ itself to use in explaining. $A(x_0)$ is the area of the intersection of the plane $x = x_0$ with $S$. This plane is parallel to the $yz$-plane, not the $xy$-plane like the $dA$ in the double integral.
The two areas are of distinct things, which is why your attempt to derive one of them from the other doesn't work. The $A$ in $dA$ is just a standard notation for integrals over two-dimensional regions. It was never intended to be the related to the cross-sectional area function $A(x)$.