The Phenomenon

The lecture I am watching contained this figure:

It is supposed to show two particles colliding in $S$ and $S’$ frames.

I wanted to check the vectors on this slide using lecture notes, but I failed.

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Context

Below I outline the context of which this problem above appears.

4-velocity basics

Relying on lecture notes:

Let’s write the proper velocity, $u^{mu}=frac{dx^{mu}}{dtau}$ as $$u^{mu}=left(cfrac{dt}{dtau},frac{dx}{dtau},frac{dy}{dtau},frac{dz}{dtau}right)=frac{dt}{dtau}left(c,frac{dx}{dt},frac{dy}{dt},frac{dz}{dt}right)=frac{dt}{dtau}(c,vec{u})$$

It is also said that proper time is the time measured by an ideal clock carried by the particle, and is related to the invariant path length by $$ds^2 = c^2 dtau^2$$

The tangent vector to the wordline is the 4-velocity of the particle, and has components: $$u^{mu}=frac{dx^{mu}}{dtau}$$

Since proper time is an affine parameter, the length-squared of the 4-velocity, ie $eta_{munu} u^{mu} u^{nu}$ is constant:

$$eta_{munu} u^{mu} u^{nu} = left(frac{ds}{dtau}right)^2 = c^2$$

Where the first equality comes from the fact that the length of the tangent vector is the derivative of the proper path length $s$ along the curve with respect to the parametrisation, here $u$. Second equality comes from $ds^2 = c^2 dtau^2$.

Using this equality we can find a relation between coordinate and proper time. We’ll need: $eta_{munu}=text{DIAG}(1,-1,-1,-1)$, so:

$$eta_{munu} u^{mu} u^{nu} = left(frac{dt}{dtau}right)^2(c^2-|vec{u}|^2)$$

We end up with:

$$frac{dt}{dtau}=left(1-frac{|vec{u}|^2}{c^2}right)^{-frac{1}{2}} = gamma_u$$

Velocity transform laws

Between $S$ & $S’$, the 4-velocity transforms are described by this equation:

$$u’^{mu}=Lambda^{mutext{ }}_{text{ }nu}u^{nu}$$

ie:

It is claimed that (and I have checked it and it is true) that from the first component of the equaition above we can have:

$$frac{gamma_u}{gamma_{u’}}=frac{1}{gamma_v}frac{1}{1-frac{vec{u}^1v}{c^2}}$$

Combine this above with the other components, get:

$$vec{u}’^1=frac{vec{u}^1-v}{1-frac{vec{u}^1v}{c^2}}$$
$$vec{u}’^2=frac{vec{u}^2}{gamma_v(1-frac{vec{u}^1v}{c^2})}$$
$$vec{u}’^3=frac{vec{u}^3}{gamma_v(1-frac{vec{u}^1v}{c^2})}$$

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My Solution

Before the collision, the left particle travels with $gamma_u(u,0)$ in $S$, which is to my understanding, is the 2-velocity (ie x and y coordinates) part of the 4-velocity. Transform this to $S’$:

To transform the first component, ie $u$ to $S’$, I’ll need the $vec{u}^1$ formula above. $vec{u}^1$, ie the x-directional velocity component of the particle in $S$ is $u$, $S’$ moves with respect to $S$ with speed $v$, standard configuration, using results above:

$$u’ = frac{u-v}{1-frac{uv}{c^2}}$$

The problem

This is not equal to $$gamma_ugamma_v(u-v)$$ as we might have hoped based on the figure.

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Question

What is going on here? Is my reasoning for the x-component of velocity in $S’$ to be $frac{u-v}{1-frac{uv}{c^2}}$ wrong, or is the figure wrong? or both the figure and my derivation is correct but I misunderstand what the figure is showing?