Let's say that a particle moves horizontally with a velocity of $v_x$ and vertically with a velocity of $v_y$. Now, when we represent these two velocities and their resultant velocity using the head to tail method, we draw a horizontal arrow which represents $v_x$ and we draw a vertical arrow whose tip is at the head of the horizontal arrow. The vertical arrow represents $v_y$.

Now, I don't know if this just happens to me but to me, the head to tail method gives the idea that the first velocity acted first and the second velocity acted once the first one was done acting but actually, both of them acted on the particle simultaneously. This just confuses me.

One approach that I have tried was that first I proved that the path that the particle follows due to two velocities acting simultaneously on it is straight. Now, we know what the position of the particle at a certain point in time will be. Let's suppose the $v_1$ represents the horizontal velocity of the particle and $v_2$ represents the vertical velocity of the particle. Then, after a time $t_1$, the horizontal displacement suffered by the particle would be $v_1t_1$ and the vertical displacement suffered by it would be $v_2t_1$. Now, let $A$ be the point that denotes the final position of the particle after time $t_1$. Let $B$ be the point that denotes the initial position of the particle (at time = $0$). Let the angle made by $Ab$ with the horizontal be $\alpha_1$. Now, $\tan\alpha_1 = \dfrac{\text{Vertical displacement in time }t_1}{\text{horizontal displacement in time }t_1} = \dfrac{v_2t_1}{v_1t_1} = \dfrac{v_2}{v_1}$.

If we do the same thing but for time equal to $t_2$, and if we take the angle made this time as $\alpha_2$, then $\tan\alpha_2 = \dfrac{v_2}{v_1} = \tan\alpha_)$

Now, $\alpha_1 \in \Big(0,\dfrac{\pi}{2}\Big)$ and $\alpha_2 \in \Big(0,\dfrac{\pi}{2}\Big)$ and $\tan\alpha_1 = \tan\alpha_2 \implies \alpha_1 - \alpha_2$.

This implies that the path that the particle follows after moving with two velocities simultaneously is straight and now we can evaluate the resultant velocity of the particle which would be the velocity of the particle along that resultant path. Now, after a time $t$, the resultant displacement of the particle $(s)$ becomes : $\sqrt{(v_1t)^2 + (v_2t)^2} = \sqrt{v_1^{~2}t^2 + v_2^{~2}t^2} = \sqrt{t^2\cdot (v_1^{~2} + v_2^{~2})} = t \sqrt{v_1^{~2} + v_2^{~2}}$

Now, the resultant velocity will be $\dfrac{s}{t} = \dfrac{t \sqrt{v_1^{~2} + v_2^{~2}}}{t} = \sqrt{v_1^{~2} + v_2^{~2}}$ which follows the head to tail rule.

Now, first of all, proving that vector quantities follow the head to tail rule for quite a few quantities will get quite tedious and length and still, I wouldn't be able to cover all vector quantities. And even if I somehow do prove this for a lot of vector quantities, I would still not be able to interpret the head to tail rule properly because it just makes it look like one of the vector quantities is acting first and once it is done acting, the second vector quantity comes into play, which is actually not the case.

So, I think that my first step would be to get a clear understanding of the head to tail rule. I want to get rid of the confusion and gain clarity on the statement that I've typed in bold.

Thanks!