Why is the shell method not
$$\lim_{n \rightarrow \infty} \sum_{k=1}^n 2\pi\left(\frac{(b-a)}{n}\right)\cdot f\left((k-1)\cdot\frac{(b-a)}{n}\right)\cdot \frac{1}{n} + \pi f\left((k-1)\cdot\frac{(b-a)}{n}\right) \cdot \left(\frac{1}{n}\right)^2.$$?
This means the shells are concentric circles going outward, the subsequent shells after the first are hollow in the center, and the hole increases further towards the end points in increasing order.
I have trouble with the implication that in the "shell method" to calculate volumes of revolution, the left end point of the first piece is $0$, the left end point of the second piece is $\dfrac{(b-a)}{n}$, the left end point of the third piece is $2\dfrac{b-a}{n}.$
Why do you use the left end point of the pieces in the integral rather than
$\int$ (constant "radius" of cylinder) $\cdot 2\pi\cdot$ height of shell $\cdot dx$?
If you're shells radius is not a constant, and the shell radius are in the increasing sequence $a_k=(k-1)\dfrac{b-a}{n}$, then you're are using overlapping shells, you're not partitioning the volume into a bunch of non-overlapping shells centered on the axis, and you're double-counting volume multiple times.
It looks like you're dealing with the case of a cylinder (i.e. the height is always $y$). In general, the height of a shell is a function of $x$, say $f(x)$, so the volume of a shell is $$\Delta V=2\pi xf(x)\Delta x + \pi f(x) (\Delta x)^2.$$ Now say the range of $x$ values is $[a,b]$. If we partition this into $n$ pieces, the left endpoint of the $k$th piece is $(k-1)\cdot \frac{(b-a)}{n}$. This is the inner radius of the $k$th shell. Now the total volume of the $k$ shells is $$\sum_{k=1}^n 2\pi\left((k-1)\cdot\frac{(b-a)}{n}\right)\cdot f\left((k-1)\cdot\frac{(b-a)}{n}\right)\cdot \frac{1}{n} + \pi f\left((k-1)\cdot\frac{(b-a)}{n}\right) \cdot \left(\frac{1}{n}\right)^2.$$
If we partition this into $n$ pieces, the left endpoint of the $k$th piece is $(k-1)\cdot \frac{(b-a)}{n}$.
To find the element of volume contained in a shell of inner radius $r = x$ and out radius $R = x + \Delta x$, length $y$, we have:
$$\begin{align*}\Delta V &= \pi(R^2-r^2)y\\ &=\pi y(x^2+2x\Delta x + \Delta x^2 - x^2)\\&=2\pi xy\Delta x + \pi y\Delta x^2\end{align*}$$
As $\Delta x$ is very small, $(\Delta x)^2$ is negligible, hence $$\Delta V = 2\pi xy\Delta x\\\therefore V = 2\pi \int_a^bxy\,dx $$
For the time being, ignore the solid of revolution and go back to the basic Riemann sum of $f(x)$ over the interval $[a,b]$ using an equally spaced partition.
We divide the interval $[a,b]$ into $n$ equal subintervals. Since the width of the interval is $b-a$, the width of each subinterval is $\frac{b-a}{n}$. On this matter, surely there can be no disagreement.
The first (leftmost) subinterval has its left endpoint at $a$. So to determine the right endpoint of the first subinterval, you must add to $a$ the width of the subinterval; i.e., the right endpoint must be $$a + \frac{b-a}{n}.$$ This is also the left endpoint of the second subinterval. So the right endpoint of the second subinterval is $$a + \frac{b-a}{n} + \frac{b-a}{n} = a + 2 \frac{b-a}{n},$$ and so on, creating an arithmetic sequence of endpoints $$a,\; a + \frac{b-a}{n},\; a + 2 \frac{b-a}{n},\; a + 3 \frac{b-a}{n}, \;\ldots,\; a + k \frac{b-a}{n},\; \ldots,\; a + n \frac{b-a}{n}$$ whose initial term is $a$, with common difference $\frac{b-a}{n}$, and last term is $a + n \frac{b-a}{n} = a + b-a = b$.
Then the (right subinterval endpoint) Riemann sum for this partition is the value of the function evaluated at each of the right endpoints, multiplied by the width of each subinterval:
$$\sum_{k=1}^n f\left(a + k \frac{b-a}{n}\right) \cdot \frac{b-a}{n},$$
and if we take the left subinterval endpoints, it looks like this:
$$\sum_{k=0}^{n-1} f\left(a + k \frac{b-a}{n}\right) \cdot \frac{b-a}{n},$$ or with a shift of the index variable $k$, like this: $$\sum_{k=1}^n f\left(a + (k-1) \frac{b-a}{n}\right) \cdot \frac{b-a}{n}.$$
But in none of these cases does it look like anything you wrote, because you have $$f\left((k-1)\frac{b-a}{n}\right).$$ When $k = 1$, my expression correctly gives either $f(a)$ (left endpoint) or $f(a + \frac{b-a}{n})$ (right endpoint), but your expression gives $$f(0).$$ This immediately disqualifies your result.
The correct way to conceptualize the cylindrical shell is to start from a correct understanding of the Riemann sum's partition structure, then take a representative subinterval and compute its volume as it is swept. If we assume that the axis of rotation is the $y$-axis and that $0 < a < b$, then a representative subinterval would have the form $$x \in \left[a + (k-1) \frac{b-a}{n}, a + k \frac{b-a}{n}\right].$$ In fact, this is the $k^{\rm th}$ subinterval for $k \in \{1, 2, \ldots, n\}$. If we take the left endpoint for the height of the cylindrical shell, this gives us $h = f\left(a + (k-1) \frac{b-a}{n}\right)$. The outer radius of the shell is the right endpoint $r_1 = a + k \frac{b-a}{n}$, and the inner radius is the left endpoint $r_2 = a + (k-1) \frac{b-a}{n}$. So its volume is exactly $$\begin{align} \pi (r_1^2 - r_2^2) h &= \pi (r_1 + r_2)(r_1 - r_2) h \\ &= \pi \left(2a + (2k-1)\frac{b-a}{n}\right)\left(\frac{b-a}{n}\right) h \\ &= 2\pi f\left(a + (k-1)\frac{b-a}{n}\right) \left(a + (k-\tfrac{1}{2}) \frac{b-a}{n}\right) \frac{b-a}{n}. \end{align}$$ However, in the limit as $n \to \infty$, it is unnecessary to take this Riemann sum: it is better to use the differential volume $$dV = 2 \pi x f(x) \, dx$$ of a representative shell with radius $x$. We can argue that its exact volume for a width of $\Delta x = \frac{b-a}{n}$ is $$\begin{align} \Delta V &= \pi \left((x + \Delta x)^2 - x^2\right) f(x) \\ &= \pi \left(2x \,\Delta x + (\Delta x)^2\right) f(x) \\ &= 2\pi x \, \Delta x f(x) + \pi f(x) (\Delta x)^2. \end{align}$$ Informally, as $\Delta x \to 0$, $\Delta x \to dx$ and $(\Delta x)^2 \to 0$; that is to say, the second term $\pi f(x) (\Delta x)^2$ vanishes much faster than the first term, and in doing so, we recover the formula for the differential volume as shown above.
Notably, the second term quantifies the error in approximating the exact volume of the cylindrical shell as an "unrolled" rectangular slab with height $f(x)$, width $2\pi x$ equal to the circumference of the shell, and thickness $\Delta x$.