You are right with your remark, that $u_0=1$ comes from the initial condition. Apperently, the numerical solution is donated by $u$, whereas the analytical solution is $y$.
The forward Euler method for $y'(t) = f(t,y(t)) $ reads
$$ u_{n+1} = u_n + h\cdot f(t_n,u_n)$$
You have $y'(t) =\lambda y(t)$, so $f(t_n,u_n)= \lambda u_n$. Now we plug this into the equation above and obtain
$$u_{n+1} = u_n + h\cdot f(t_n,u_n) = u_n+ h\lambda u_n = u_n\cdot (1+h\lambda)$$
But we can furthermore express $u_n$ in terms of $u_{n-1}$ and so on, up to $u_0$
\begin{align}u_{n+1} &= u_n\cdot (1+h\lambda)
\\ &=u_{n-1}\cdot (1+h\lambda)\cdot (1+h\lambda)
\\ &= ...
\\ &=u_0\cdot (1+h\lambda)\cdot (1+h\lambda) \cdot ...\cdot (1+h\lambda)
\\ &=u_0 (1+h\lambda)^n = (1+h\lambda)^n\end{align}
Since $u_0=1$.
Now assuming you mean $\lambda <0$ which makes sense, because otherwise your solution would diverge and that would be a bad model problem.
Furthermore I assume you mean $ -1<(1+\lambda h) <1 $ iff $ h < 2/|\lambda|$ and not
$ -1<(\lambda h) <1 $ iff $ h < 2/|\lambda|$ as you want to show that your solution reaches zero if $n\rightarrow \infty$.
So to get $1+h\lambda<1$ you get $h\lambda < 0 \Rightarrow h>0$ as $\lambda <0$.
For $-1<1+h\lambda$ you now have $-2<h \lambda \Rightarrow -2/\lambda > h$. With $\lambda<0$ you get $-2/\lambda = 2/|\lambda|$ So finally you have $0<h<2/|\lambda|$.
Please let me know, if my assumptions ($\lambda<0$ and the different inequality) were correct.
Best Answer
The function $Q$, which more often than not will be a polynomial or rational function, is the factor that approximates the exponential $e^{hκ}$ in the numerical solution of $w'(t)= κw(t)$. In the exact solution $w(t_{i+1})=e^{hκ}w(t_i)$, in the numerical solution $w_{i+1}=Q(hκ)w_i$. In your scheme \begin{align} w^*_{i+1}&=w_i+h·(κw_i)=(1+hκ)w_i \\ w_{i+1}&=w_i+\frac h2 ((κw_i)+κ(1+hκ)w_i)=(1+hκ+\tfrac12(hκ)^2)w_i \end{align}