Just to be clear, you highlighted your exact problem when you wrote the question in “length of train for someone on the train (¿ but from my point of view).” Τhere is a mixed perspective happening here.
You have correctly defined a $\Delta t$ as the time between two objective events, one of them being the front of the train passing you, and one of them being the back of the train passing you. You have correctly calculated that during this time interval if you are watching someone in the train holding a clock, their clock only increments by $\Delta t/\gamma$ between these two events. And you have correctly calculated that you think the length of the train is $v\Delta t$ but now you incorrectly transfer to the train’s perspective.
Why is this incorrect? Because of the relativity of simultaneity.
Put a clock at the front and back of the train, that the person in the train thinks are synchronized. You, on the ground, do not think they are synchronized! Rather you see them both tick at $1/\gamma$ seconds per second and also the one at the front of the train always seems to be earlier than the one at the back of the train by a fixed offset $\sigma$. And here we have to be a little bit clear, $\sigma$ is just the difference between the two passing numbers on the clocks, at a time that I on-the-ground think is simulataneous.
To understand what the person in the train thinks happened, we can then just write down the times of these two clocks as they pass you. So the one at the front initially shows $t'$, then the one at the back shows $t' + (\Delta t/\gamma) + \sigma,$ and so according to the person in the train, the time it took you to pass by their entire train was $(\Delta t/\gamma) + \sigma,$ because they think these clocks are synchronized... and their calculation of the length of the train is therefore longer than you thought, by a distance $v~\sigma$.
What is this offset $\sigma$ more concretely? It is $L_0 v/c^2$ where $L_0$ is the length that the person on the train thinks the train is.[1]
If you are thinking like an ordinary person you will now say, “But, that's what I wanted to calculate! You can't tell me that to derive $L_0$ I first have to know $L_0$!”. But we think like physicists and do this all the time, just set up the equation:
$$
\begin{align}
L_0 &= v{\Delta t\over\gamma} + v\sigma\text{, where } \sigma = \frac{vL_0}{c^2},\\
L_0 &= v{\Delta t\over\gamma}
+L_0\frac{v^2}{c^2},\\
&\text{(collect } L_0\text{ terms on the left})\\
\frac{L_0}{\gamma^2} &=v{\Delta t\over\gamma}\\
L_0&=\gamma v \Delta t,\end{align}$$
which was to be proven.
The technical term for thinking this way is to say that there is only one “self-consistent solution” or so.
Note 1: This math can get a bit subtle. Don't get caught up putting a $\gamma$ in this expression $L_0v/c^2$ by using the Lorentz transform directly, because the land-time that you transform the train-time zero to will not be $t=0$ and that messes everything up. Instead you want to Lorentz-boost the land-coordinates $(t, x) = (t, vt-L)$ into the train frame to get $(t/\gamma +\gamma L v/c^2, \gamma L)$, versus boosting just $(t, vt)$ at the front of the train to get $(t/\gamma, 0)$. The distance between these lines at constant time is now super easy: the rest length is the de-contracted $L_0=\gamma L$, while at $t=0$ the clock at the back of the train must have been showing $t'=+L_0 v/c^2,$ which is our $\sigma.$ (Of course if you do it this way then you already prove the length contraction formula en route to providing the fact $\sigma=L_0v/c^2$ that explains the answer... c'est la vie.)
