Suppose i have two curves $\gamma_1$ and $\gamma_2$, where one is a linear curve, $$\gamma_1(t)=tP_1+(1-t)P_2$$ and the other is a similar path, composed with a $C^\infty$-diffeomorphism $\Phi:\mathbb{R}^n\rightarrow\mathbb{R}^n$: $$\gamma_2(t)=\Phi(tP_3+(1-t)P_4)$$
Can one prove, that the intersections are finite closed intervals, where I consider $[a,a] = {a} \in \mathbb{R}^n$ a closed interval as well?
I think it's wrong. First consider the curve $$ \gamma(t) = \begin{cases} (t,0) & \text{ if } |t| \geq 1 \\ (t, e^{-\frac{1}{1-t^2}}) & \text{ if } |t| < 1 \end{cases} $$ It is $C^{\infty}$ and agrees with the curve $\alpha(t) = (t,0)$ on two disjoint unbounded intervals. Since you want $\gamma$ to be the image curve under a diffeomorphism I suggest the diffeomorphism $$ \phi : (x,y) \rightarrow \begin{cases} (x,y) & \text{ if } |x| \geq 1 \\ \left(x,y+e^{-\frac{1}{1-x^2}} \right) & \text{ if } |x| < 1 \end{cases}. $$ Which has the inverse $$ \phi^{-1} : (u,v) \rightarrow \begin{cases} (u,v) & \text{ if } |u| \geq 1 \\ \left(u,v-e^{-\frac{1}{1-u^2}} \right) & \text{ if } |u| < 1 \end{cases}. $$ And consider the curve $\phi(t,0)$ which gives you $\gamma$.
EDIT:
Define the intersection set $$ X=\{ (s,t) \in ]a,b[ \times [0,1] | \alpha(s) = \gamma(t) \}. $$ Firstly it is closed:
Take $(s_n,t_n) \rightarrow (s,t)$ a converging sequence in $X$. Then $$ \alpha(s) = \lim_n \alpha_n(s_n) = \lim_n \gamma(t_n) = \gamma(t). $$ This shows $(s,t) \in X$, hence $X$ is closed.
Secondly the values $t$ can take is clearly bounded since it is a subset of $[0,1]$.
Lastly the values $s$ can take is bounded since if we write $\alpha = (\alpha_1,\alpha_2)$ we have $$ X = \{ (s,t) \in ]a,b[ \times [0,1] | \alpha_1(s) = \gamma_1(t) \} \cap \{ (s,t) \in ]a,b[ \times [0,1] | \alpha_2(s) = \gamma_2(t) \}. $$ Now if $\alpha$ is not a point, then at one of the $\alpha_i$'s is invertible. Assume w.l.o.g. we can take $\alpha_1$. Write $\alpha_1(s) = zt+w$. Then $$ \{ (s,t) \in ]a,b[ \times [0,1] | \alpha_1(s) = \gamma_1(t) \} = \{ (s,t) \in ]a,b[ \times [0,1] | s = \gamma_1(t)/z-w/z \}.$$ The values $s$ achieves in this set is then given by the continuous image of $\gamma_1(t)/z-w/z$ under $[0,1]$. Hence it is compact and hence bounded.
So $X$ is written as the intersection of two sets of which one is certainly bounded. Combining this with the closedness of $X$, we conclude it is compact.