Consider the formula $1 + \frac{y}2$. This has a fixed point at $y = 2$. And if we use the equation $y = 1 + \frac{y}2$ to substitute for $y$ in our formula, we get $1 + \frac{1 + \frac{y}2}2$, or $1 + \frac12 + \frac{y}4$. If we keep repeating we get $1+\frac12 + \frac14 + \dots$, which does indeed equal our fixed point $2$.
But now consider the formula $1+2y$. This has a fixed point at $y=-1$. Here if we use the equation $y = 1+2y$ to substitute for $y$ in our formula, we get $1+2(1+2y)$, or $1+2+4y$. If we keep repeating we get $1+2+4+\dots$, which does not equal our fixed point $-1$.
Can someone explain in simple terms why this procedure works in the first case but not the second? Would the procedure work in every case where the generated series converged?
Aside from the argument that $1+2+4+\cdots$ could equal $-1$, to answer your question, there are two classes of fixed points.
An attractive fixed point is a fixed point where within a certain range, a point fed into the formula recursively will eventually converge to the fixed point. Your first example is an attractive fixed point.
A fixed point that is not attractive has no points converging to the fixed point when fed into the formula recursively. Your second example is a non-attractive fixed point.
The property of attractive fixed points is that the derivative (i.e., slope) of the function in the formula has magnitude less than $1$ on the range of points that converges to the fixed point:
$$|f'(x)|<1$$
where $x_{n+1}=f(x_n)$ is the recursive formula.
See this (fixed point) and this (fixed-point iteration) for more information and a clearer explanation.