I am about to show this well-known theorem that states If $y\in Y$ is a regular value of $f: X\rightarrow Y$, then $f^{-1}(\{y\})$ is a closed submanifold of $X$ and furthermore the identity $T_xf^{-1}(\{y\}) = kerT_xf$ holds. I was able to show it is indeed a submanifold and for the second part I found a proof of which I do not understand one specific aspect.
Suppose we have shown the first part and I will write down the proof of the second part. Let $x\in f^{-1}/\{y\}), y\in T_xX$. Since $v$ is in the vectorspace containing $\phi(U)$ where $(\phi,U)$ is a chart of $x$ and $(\psi,V)$ is a chart of $f(x)=y$ and because $y$ is regular value of $f$, and so $f$ Submersion at $x$, we have $\phi(U)= U_1 \times U_2$, and by that we write $v=(v_1,v_2)$. On the other hand, because $f$ is Submersion at $x$, we know by definition that $$ f_{UV}: \psi\circ f\circ \phi^{-1} $$ , the map from the definition of differentiability of a function between manifolds, can be written as $U_1\times U_2 \xrightarrow{pr_1} U_1 \xrightarrow{\sim} \psi(V)$. Furthermore, we can identify $T_xfv$ as $Df_{UV}(\phi(x))v$, by definition of $T_xf:T_xX\rightarrow T_{f(x)}Y$.
We now observe the following identity: $Df_{UV}(\phi(x)) \overset{(1)}{=} d/dt f_{UV}(\phi(x)+tv)|_{t=0}$
where (1) is the chain rule. Splitting the argument into its coordinates and having $U_1=0$ because of submanifold property, we remember the factorisation of $f_{UV}$ that gives
$d/dt f_{UV}(\phi(x)+tv) |_{t=0} =d/dt tv_1 |_{t=0}=v_1$
and this is the aspect I do not understand. Because with the factorisation we get
$f_{UV}$ firstly maps the argument onto $pr_1(\phi(x)+tv_1)$ with $pr_1(\phi(x))\in U_1=0$. But then comes the diffeomorphism and by looking at the proof, the diffeomorphism leaves $tv_1$ untouched, but why?
Having this identity, with the first remarks done above, one can quickly prove the set inclusions.