Definitions
L
$L$ is not the length of the interval, but a Lipschitz constant: some
number such that $$|f(x,y_1) - f(x,y_2)| \le L|y_1 - y_2|$$
satisfied for all $y_1,y_2,x$ (in some interval of interest).
M
Similarly, $M$ is any number such that the assumption $|y''(x)| \le M$
is satisfied. The error analysis is also explained in this article
here (reference taken from the Wikipedia article on the Euler
method).
Example
Suppose the equation $y' = -\frac12 y$ with initial condition $y(0) = 1$ with the Euler method on the interval $x \in [0,1]$. We actually know the exact solution in this case, $y(x) = \exp(\frac12x)$.
Finding L
We note that $f(x,y) = \frac12 y$ in the example, so the equation
$$|f(x,y_1) - f(x,y_2)| \le L|y_1 - y_2|$$
becomes
$$|\frac12 y_1 - \frac12 y_2| \le L |y_1 - y_2|.$$
This equation is satisfied for $L = \frac12$, so this is the value of $L$ we will take (we can also have taken a bigger value for $L$, like $L=1$, in which case we'll get a result which is also valid but not as sharp).
Finding M
We need it to satisfy $|y''(x)| \le M$. Since we know the exact solution, $y(x) = \exp(\frac12x)$, we know that $y''(x) = \frac14\exp(\frac12x)$ and so $|y''(x)| \le \frac14$ on the interval $x \in [0,1]$. However, for most equations we do not know the exact solution, so let us pretend we do not. Then what we can do is the following: differentiate the differential equation $y'(x) = \frac12 y$ to get $y''(x) = \frac12 y'$, and then substitute the differential equation in it to get
$$y''(x) = \frac12 \cdot \frac12 y = \frac14y.$$
We do know that the numerical solution given by the Euler method is a decreasing sequence for this example, so $y\le1$ and thus $|y''(x)| \le \frac14$ (as we found before).
Conclusion
So, we can take $L = \frac12$ and $M = \frac14$ in the error bound.
First, we get local truncation error.
$x(t_{k+1}) = x(t_k) + hf(t_k,x(t_k)) + \tau_k$
$\tau_k = x(t_{k+1}) - x(t_k) - hf(t_k,x(t_k)) = \frac{h^2}{2}x''(\eta)$. Where $\eta \in (t_k,t_{k+1})$.
Then we get the bound,
$$|x(t_{k+1}) - x_{k+1}| \le (1+hL)|x(t_{k}) - x_{k}| + |\tau_k|$$ $$\le (1+hL)|x(t_{k}) - x_{k}| + \frac{h^2}{2}\max_{s \in (0,T)}|x''(s)|$$ $$\le \frac{1}{1-hL}|x(t_{k}) - x_{k}| + \frac{h^2}{2}\max_{s \in (0,T)}|x''(s)|$$
Where the last step is from geometric series. Maybe it would be helpful if you listed the other results you are talking about, and then we can show that they're equivalent.
Best Answer
EDIT:
The simple argument contains an error which is revealed by studying the initial value problem \begin{align} x(0) &= 1 \\ x'(t) &= x(t) \end{align} In this case, all quantities can be computed explicitly and compared. The exact solution is $$x(t) = e^t.$$ Euler's method takes the form \begin{align} y_0 &= 1 \\ y_{j+1} &= y_j + y_j h. \end{align} It follows that $$y_n = (1+h)^n.$$ The truncation error at $t_j = jh$ is $$ \epsilon_j = x(t_{j+1}) - \left( x(t_j) + x(t_j)h\right) = e^{(j+1)h} - e^{jh}(1+h).$$ It follows that the sum of the truncation errors are $$ T_n = \sum_{j=0}^{n-1} \epsilon_j = e^h \sum_{j=0}^{n-1} (e^h)^j - (1+h) \sum_{j=0}^{n-1} (e^h)^j = (e^h - 1 - h) \frac{e^{nh}-1}{e^h-1}.$$ In contrast, the global error at $t=nh$ is $$ E_n = e^{nh} - (1+h)^n.$$ We have $T_n = E_n$ if and only if $$ (e^{nh} - (1+h)^n)(e^h - 1) = (e^h - 1 - h)(e^{nh}-1)$$ or equivalently $$ e^{(n+1)h} - e^{nh} - (1+h)^n e^h + (1+h)^n = e^{(n+1)h} - e^h - e^{nh} + 1-he^{nh} + h$$ or equivalently $$(1-e^h)(1+h)^n = 1-e^{h} + h(1-e^{nh})$$ or equivalently $$(1+h)^n = 1 + h \frac{e^{nh} - 1}{e^h - 1}$$ or equivalently $$\frac{(1+h)^n - 1}{h} = \frac{e^{nh} - 1}{e^h - 1}$$ Now for fixed $n$ the left hand side is a polynomial in $h>0$ of degree $n-1$. The right hand side is not a polynomial of $h>0$ of degree $n-1$ unless $n=1$. We conclude that $T_n = E_n$ if and only if $n=1$.
The simple argument contains an error which hinges on the difference between the global error and the local truncation error.
In order to eliminate any doubt, I will give all relevant definitions Consider the initial value problem \begin{align} x'(t) &= f(t,x(t)) \\ x(t_0) &= x_0 \end{align} where $f$ is such that there exists a unique solution $x$ for any initial value $(t_0,x_0)$. Let $h > 0$ be given and set $t_j = t_0 + jh$. Euler's explicit method is given by \begin{align} y_{j+1} &= y_j + h f(t_j,y_j), \quad j = 0,1,2,\dots \\ y_0 &= x_0 \end{align} and we use $y_j$ as an approximation of $x(t_j)$. The global error $e_j$ is given by $$e_j = x(t_j) - y_j, \quad j = 0,1,2\dotsc $$ In order to bound the global error we need an error formula. We have an iteration for $y_j$, and so it is natural to seek an iteration for $x_j:=x(t_j)$. By Taylor's formula we have at least one $\xi_j \in (t_j,t_{j+1})$, such that $$x_{j+1} = x(t_{j+1}) = x(t_j + h) = x(t_j) + x'(t_j)h + \frac{1}{2}x''(\xi) h^2 \\= x_j + f(t_j,x_j)h + \frac{1}{2}x''(\xi_j) h^2.$$ We conclude that the $x_j$ satisfy an iteration which is quite similar to Euler's method. The local truncation error $\epsilon_j$ at time $t_{j}$ is the error term given by $$\epsilon_j = \frac{1}{2}x''(\xi_j) h^2.$$ We see that the local truncation error is the difference between $x(t_{j+1})$ and a single step of Euler's method starting from the initial point $(t_{j},x(t_j))$, i.e., $$ x_{j+1} - ( x_j + f(t_j,x_j)h) = \frac{1}{2} x''(\xi_j) h^2.$$ It is this relationship which allows us to derive an inequality of the type $$ |e_{j+1}| \leq (1 + Lh) |e_j| + \frac{1}{2} M h^2,$$ which leads to the bound $$ |e_n| \leq \frac{1}{2} Mh \frac{e^{LT} - 1}{L}, \quad 0 \leq nh \leq T$$ However, in general it is not true that the global error satisfies $$e_n = \epsilon_0 + \epsilon_1 + \epsilon_2 + \dotsc + \epsilon_{n-1}$$ The right hand side incorporates information about single Euler steps from $n$ different points on the exact solution curve. In contrast to the left hand side, it does not have information about the repeated application of Euler's method starting from a single point.