This is a proof of the invertibility of the matrix only, when the $\lambda_i$'s are distinct.
Let $a_0,a_1,\ldots,a_{n-1}$ be coefficients, such that the corresponding linear combination of the rows of $V_{m_1,\ldots,m_k}(\lambda_1,\ldots,\lambda_k)$ is zero.
Denote by $P$ the following polynomial:
$$P=a_0+a_1x+\cdots+a_{n-1}x^{n-1}$$
The assumption is equivalent to the following equalities holding true:
$\frac{P(\lambda_1)}{0!}=0,\frac{P'(\lambda_1)}{1!}=0,\ldots,\frac{P^{(m_1-1)}(\lambda_1)}{(m_1-1)!}=0$
$\frac{P(\lambda_2)}{0!}=0,\frac{P'(\lambda_2)}{1!}=0,\ldots,\frac{P^{(m_2-1)}(\lambda_1)}{(m_2-1)!}=0$
$\ldots$
This implies that each of the following polynomials divides $P$: $(x-\lambda_1)^{m_1}$, $(x-\lambda_2)^{m_2}$, $\ldots$ , $(x-\lambda_k)^{m_k}$.
As the $\lambda_i$ are distinct, we get: $P(x)=Q(x)\prod_{p=1}^{k}(x-\lambda_p)^{m_p}$.
As $P$ is of degree less than $n$, this gives $Q=0$, so $P=0$, so all the $a_i$'s are zero.
Hence, the matrix $V_{m_1,\ldots,m_k}(\lambda_1,\ldots,\lambda_k)$ has linearly independent rows; it's invertible.
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
Hint: Consider $B = E_1 E_2 \ldots E_{i-1} A$ where $E_k$ is the elementary matrix associated with swapping the $k$-th row and the $(k+1)$-th row. Recall the determinant of a product of matrices is the product of the determinants, and elementary matrix w.r.t. type I ERO has determinant $-1$.