Usually, to analyze a simple pendulum problem we look at the equation
$$ma = mg\sin(\theta)\tag{1}$$
where $\theta \in [0,\pi/2]$ is the variable that shows the angle between the string of the pendulum and a vertical axis for the pendulum at rest. See figure below.
And with calculus we have that this equation can be put in the form
$$m\ddot\theta L = mg \sin(\theta) \implies \ddot \theta -\frac{g}{L}\sin(\theta) = 0$$
And we solve for angles $\theta << 1$ such that this equation is solvable with solution $\theta(t) = \theta_0\cos(\omega t+\delta_0)$ where $\omega = \sqrt{g/L}$ is the angular velocity. But instead of using calculus I was trying to show this fact using just geometry and the $\theta << 1$ approximation.
Using that $a$ in $(1)$ is the centripetal acceleration, then we get that, setting the approximation $\sin(\theta) = \theta$
$$a = g\theta = \frac{gs}{L} = \frac{v^2}{L} = \omega^2L \implies \omega^2 = \frac{gs}{L^2} \tag{2}$$
Where I just used that the centripetal acceleration is $a = v^2/L$ and that $\omega L = v$. Now we look at the triangle below.
Using the cosines law we have that
$$\tilde{L}^2 = 2L^2 - 2L^2(\cos(2\theta)) = 2L^2 - 2L^2(\cos^2(\theta)-\sin^2(\theta)) \stackrel{\theta<<1}{=} 2L^2\theta^2 = 2s^2$$
so we get that $s = \tilde{L}/\sqrt{2}$ such that $(2)$ becomes
$$\omega^2 = \frac{g}{L^2}\cdot \frac{\tilde{L}}{\sqrt{2}}\tag{3}$$
Now we note that we have that $h = \sqrt{L^2-\tilde{L}^2/4}$ and that $\tan(\theta) \stackrel{\theta<<1}{=} \theta $ and then we get
$$\frac{\tilde{L}/2}{h} = \frac{s}{L} = \frac{\tilde{L}}{\sqrt{2}L}$$
but then
$$\color{blue}{h = \frac{\sqrt{2}L}{2} } \implies L^2-\frac{\tilde{L}^2}{4} = \frac{L^2}{2} \implies \tilde{L} = \sqrt{2}L \tag{4}$$
And using $(4)$ in $(3)$ we get
$$\omega^2 = \frac{g}{L}\tag{5}$$
Wich is the result wanted.
My question is:Is there any problem using these arguments to obtain the equation $(5)$? I'm a bit confuse with the blue equation, because $h$ seems to be constant because $L$ is constant but as the angle approach zero this shouldn't be the case. The main goal is to show the result without using calculus. These approximations should be done carefully? And if yes, why and how?
You are confusing the centripetal with tangential and resultant, or net, acceleration. Newton's law says that $$ m{\bf a} = {\bf F}_{net} = m{\bf g} + \bf T$$ and here we have two different forces acting on the object - gravitational and tension (as in your picture). $\bf T$ is always along the rope, so points towards the center of the circular trajectory, and contributes only to centripetal acceleration. $m\bf g$ is straight down, and when resolved into components along $\bf v$ and perpendicular to $\bf v$, contributes $mg\sin\theta$ to tangential and $-mg\cos\theta$ to centripetal (minus because it pulls "out"). So $$ a_{\bf n} = T/m - g \cos \theta $$ $$ a_{\bf t} = g \sin\theta $$
The first of these is related to speed, and through that to angular velocity, while the second is related to the angular acceleration.
The misconception you have that makes you say that $T = mg\cos\theta$ comes from doing static problems. You are thinking these two forces cancel out because there is no motion in that direction. Wrong. It is not motion that determines if forces should cancel out, but acceleration. And here there is acceleration towards the center - exactly what we call centripetal.
The tension force is one of those that can adjust its magnitude to whatever is needed to accomplish a particular purpose: keep the rope intact. To do that, it has to keep the object moving in a circle, and that requires centripetal acceleration.
In many circular motion problems, the speed is constant, which means tangential acceleration is zero, and so the net is equal to centripetal. But it is not true in general, and it is not true here.
EDIT: I just noticed another major flaw in your exposition. You are using $\omega$ in two different senses. Never use the same symbol for different quantities!
The $\omega$ in $\omega L = v$ is the angular velocity of the object at some moment in time. It is simply $\omega(t) = \dot\theta$, and it is not a constant. It changes over time, just as $v$ does.
Then you call $\omega$ the thing you wanted to derive in the first place. That is called angular frequency and it is a constant. It is the parameter in all of the functions describing the motion - such as $\theta(t)$ - that relates to their periodicity.
ANOTHER EDIT: $$ \cos^2\theta - \sin^2\theta \neq 1 - \theta^2 $$ for small angles. You neglect some important terms in the cosine that way. $$ \cos^2\theta \approx (1 - \theta^2/2)^2 \approx 1 - \theta^2 $$ so the whole thing becomes $$ \cos^2\theta - \sin^2\theta \approx 1 - 2\theta^2 $$
You should then get that $\overline L \approx 2s$ as would makes sense, because for small angles the arc is almost straight.
THIRD EDIT: Sorry to keep flogging that horse, but it kept bugging me how that non-constant $\omega$ suddenly switched to the constant one, which shouldn't work out at all. So I noticed another, very subtle error: you write at first $a = g\theta = gs/L$ (and yes, it has already been pointed out how that the physics of that is wrong), but the $\theta$ here in the first equality should be the position of the object at some moment in time, i.e. it is a function of time, not a constant. In the second equality, you are thinking of it as the angle between the vertical and the endpoint, or what a physicist calls the amplitude (which is a constant). So there are two of those using-same-symbol-for-different-things mistakes here and they sort of cancel out! (Though not in the sense that the result could still be correct.)