About the last point of you question I think it's not really simple to state a closed simple formula for the determinants. I tried to see if there are symmetries. There are but not really useful (at least as far as I can see).
Just to have an idea the first 3 steps of the induction you have the following determinants.
If $n=2$ then
\begin{equation}
\det A^{(2)} = a_0^2 - a_1^2
\end{equation}
If $n=3$ then
\begin{equation}
\det A^{(3)} = (a_0 - a_2) (a_0^2 - 2 a_1^2 + a_0 a_2)
\end{equation}
Already at $n=4$ the formula is not so simple. Indeed if we have $n=4$ we the determinant become:
\begin{equation}
\det A^{(4)} = a_0^4 + a_1^4 + a_2^4 - 2 a_1 a_2^2 a_3 +4 a_0 a_1^2 a_2 - 3 a_1^2a_0^2 -2 a_2^2a_1^2 - 2 a_2^2a_0^2 + a_3^2a_1^2 - a_3^2a_0^2 - 2 a_1^3 a_3 + 4 a_0 a_1 a_2 a_3
\end{equation}
What happens is that when you have to calculate $\det A^{(n)}$ the minor determinants of order $n-1$ are matrices where the condition of symmetry that allowed huge simplifications disappears. With n=2,3 the problem is not so big since the minors are trivial, but when n gets bigger the problems arise. In fact the minors are not really Toeplitz matrix, but "block Toeplitz Matriz" (sort of saying). So maybe there could be a way of enclosing the writing in a simple notation formula, but it wouldn't be a real computational gain...
For aesthetic considerations suggested by Michael Hoppe, I'll rename $K$ into $n$ and $y_{K+1}$ into $x_{n+1}$, so the matrix whose determinant you're searching is
$$A=\begin{pmatrix}
x_1 & 0 & \dots & 0 & y_1 \\
0 & x_2 & \dots & 0 & y_2 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & x_n & y_n \\
y_1 & y_2 & \dots & y_n & x_{n+1}
\end{pmatrix}.
$$
First method. Use determinant expansion with respect to the last column to get
$$\mathrm{det}(A) = x_1...x_n x_{n+1} + \sum_{i=1}^n(-1)^{n+1+i} \mathrm{det}(A_i)$$
where $A_i$ is the matrix $A$ deprived of its last column and $i$-th row. For example,
$$A_1 = \begin{pmatrix}0& x_2 &0& ... &0 \\
0 & 0 & x_3 & ... & 0 \\
\vdots & & & \ddots& \vdots\\
0 & & ... & & x_n\\
y_1 & &... && y_n \\
\end{pmatrix}. $$
Using row expansion for $A_i$, it is easy to see that $\mathrm{det}(A_i)$ is equal to $(-1)^{n+i} y_i \prod_{j \neq i} x_j$, which yelds
\begin{align*}\mathrm{det}(A) &= \prod_{i=1}^{n+1} x_i - \sum_{i=1}^n y_i^2 \prod_{j \neq i, j\leqslant n}x_j \\
&= \prod_{i=1}^{n+1}x_i \left( 1 - \sum_{i=1}^n \frac{y_i^2}{x_i}\right).
\end{align*}
Edit (second method, hint). This last expression suggests another method (maybe it's not working) : suppose that no $x_i$ is zero. Note $X = \mathrm{diag}(x_1, ..., x_{n+1})$ and $Y = A - X$, so that $A = X+Y = X(\mathrm{Id}+X^{-1}Y)$. Then, $\mathrm{det}(A) = \mathrm{det}(X) \mathrm{det}(\mathrm{Id} - X^{-1}Y)$. Now, all you have to do is to compute $\mathrm{det}(\mathrm{Id} - X^{-1}Y)$. Maybe there's a simple ay of doing this (I don't know).
Best Answer
You matrix $A = (a_{ij})$ is an upper triangular matrix ( $a_{ij} = 0$ whenever $i > j$ ) and a Toeplitz matrix ( $a_{ij}$ depends only on $i-j$ ) at the same time. I cannot find any reference online which teach you how to evaluate its inverse effectively. I hope this keywords can help you in your own search.
If you just want an inverse without too many other concerns, it is actually pretty easy to get the inverse ourselves. Let $\eta$ be the $n \times n$ matrix with $1$ on its superdiagonal and $0$ otherwise. i.e.
$$\eta = (\eta_{ij}),\quad \eta_{ij} = \begin{cases}1,& i - j = -1\\0,& \text{otherwise}\end{cases}$$
We have $\eta^n = 0$ and we can express $A$ as a polynomial in $\eta$.
$$A = x_1 I + x_2 \eta + x_3 \eta^2 + \cdots + x_n \eta^{n-1}$$
$A$ will be invertible when and only when $x_1$ is non-zero. When $A$ is invertible, $A^{-1}$ is also an upper triangular Toeplitz matrix. We can also represent it as a polynomial in $\eta$.
Introduce numbers $\displaystyle\;\alpha_i = \frac{x_{i+1}}{x_1}$ and $\beta_i$ ( $i = 1,\ldots,n-1$ ) such that
$$\begin{align} A &= x_1 \left(I + \alpha_1 \eta + \alpha_2 \eta^2 + \cdots + \alpha_{n-1} \eta^{n-1}\right)\\ A^{-1} &= x_1^{-1} \left(I + \beta_1 \eta + \beta_2 \eta^2 + \cdots + \beta_{n-1} \eta^{n-1}\right) \end{align} $$ The condition $A^{-1} A = I$ can be expanded to following set of relations. They will alow you to compute $\beta_k$ in a recursive manner.
$$\begin{align} -\beta_1 &= \alpha_1\\ -\beta_2 &= \alpha_1 \beta_1 + \alpha_2\\ &\;\vdots\\ -\beta_k &= \alpha_1 \beta_{k-1} + \alpha_2 \beta_{k-2} + \cdots + \alpha_k\\ &\;\vdots \end{align} $$
When $n$ is small and you want individual $\beta_k$ as a function of $\alpha_k$. There is actually a trick to get it. You can ask a CAS to compute the Taylor expansion of the reciprocal of following polynomial in $t$:
$$\frac{1}{1 + \alpha_1 t + \alpha_2 t^2 + \cdots + \alpha_{n-1} t^{n-1}} = 1 + \beta_1 t + \beta_2 t^2 + \cdots + \beta_{n-1} t^{n-1} + O(t^n)$$
The coefficients of $t^k$ ($1 \le k < n$) in the resulting Taylor expansion will be the expression you want for $\beta_k$. e.g.
$$\begin{align} \beta_1 &= -\alpha_1,\\ \beta_2 &= \alpha_1^2 - \alpha_2,\\ \beta_3 &= -\alpha_1^3 + 2\alpha_1\alpha_2 - \alpha_3,\\ \beta_4 &= \alpha_1^4 - 3\alpha_1^2\alpha_2 + \alpha_2^2 + 2\alpha_1 \alpha_3 - \alpha_4\\ &\;\vdots \end{align}$$