Monty hall problem probability 2/6?

It's simply because your diagram assumes you stay with probability $1/2$. In that case the probability of switching and getting the car is indeed $2/6$, and the probability of switching and not getting the car is $1/6$.

But if you always switch, you're twice as likely to get each of these outcomes - instead of a $1/2$ probability of switching, you have probability $1$ of switching - so they become $2/3$ and $1/3$ respectively.

That indicates that instead of using a strategy, you are randomly choosing whether to swap or stay.   The probability of winning a car by doing that is and unsurprising $1/2$.

$$\tfrac 13\tfrac 12+\tfrac 23\tfrac 12=\tfrac 12$$

The point of the scenario is that if you use strategy of always swapping, the probability of winning a car becomes: $2/3$.

$$\tfrac 13\tfrac 01+\tfrac 23\tfrac 11=\tfrac 23$$

You're calculating the probability wrong. If you decide at random with equal probability to swap or stay, in $\tfrac{2}{6}$ of all cases you will have swapped and gotten the car.

But that is not the probability of if you decide to swap, would you get the car?

Note that in exactly $\tfrac{3}{6}=\tfrac{1}{2}$ of all cases, you will have swapped, and in $\tfrac{1}{2}$ of all cases you will have stayed. But look at those cases.

Of the $\tfrac{3}{6}$ cases where you will have swapped, there is a $\tfrac{2}{6}$ chance of having swapped and gotten a car while there is a $\tfrac{1}{6}$ of having swapped and gotten a goat sheep.

On the other hand there is a $\tfrac{1}{6}$ chance of having stayed and gotten a car while there is a $\tfrac{2}{6}$ of having swapped and gotten a sheep.

So if you swap, the probability of getting a car is twice that of getting a sheep, while if you stay, the chance of getting a sheep is twice that of getting a car.
Which is the exact outcome of the canonical Monty Hall problem.