The statement (of the original question) is not true. Take $v_1=e^{2\pi i/3}=(-1/2,\sqrt{3}/2)$, $v_2=e^{4\pi i/3}=(-1/2,-\sqrt{3}/2)$, $w_0=w_1=(1,0)=e^0$.
Then
- $v_1$ and $w_1$ are linearly independent
- $v_2$ and $w_2$ are linearly independent
- the following inequalities are satisfied:
$$\displaystyle{\frac{||v_2||}{||v_1||}=1\ge 1= \frac{||w_2||}{||w_1||}}\quad \text{and}\quad \frac{||v_2||}{||v_1||}=1\ge 1=\frac{||v_2+w_2||}{||v_1+w_1||}$$
For all $t$, we have $w_t=(1,0)$. For $t=1/4$ we have $v_t=e^{i((1-t)2\pi/3+t4\pi/3)}=e^{5\pi/6}=(-\sqrt{3}/2,1/2)$. Hence $v_t+w_t=(1-\sqrt{3}/2,1/2)$ and so $\|v_t+w_t\|=\sqrt{2-\sqrt{3}}<1$. Set $t_1=1$ and $t_0=1/4$. Then
$$
\frac{||v_{t_1}||}{||v_{t_0}||}=1<\frac{1}{\sqrt{2-\sqrt{3}}}=\frac{||v_{t_1}+w_{t_1}||}{||v_{t_0}+w_{t_0}||}
$$
I don't see any conditions you can put on $v_1,v_2,w_1,w_2$ such that the inequality is satisfied. What I did was to find the vectors such that you have equality for $t=0$ and for $t=1$. Then there must be an intermediate value where your inequality fails.
${\bf EDIT:}$
As for the second version of the question we also find a counterexample, that is, $v_0(v_{0x},0),v_1=(v_{1x},0),w_0,w_1$ such that
$v_{0x}>0$ and $v_{1x}>0$
$0<\theta_0<\pi$ and $0<\theta_1<\pi$
the following inequalities are satisfied:
$$\displaystyle{\frac{||v_1||}{||v_0||}> \frac{||w_1||}{||w_0||}}\quad \textit{and }\frac{||v_1||}{||v_0||}> \frac{||v_1+w_1||}{||v_0+w_0||}$$
and such that for $t_1=1/2$ and $t_0=0$ we have
$$\displaystyle{\frac{||v_{t_1}||}{||v_{t_0}||}< \frac{||v_{t_1}+w_{t_1}||}{||v_{t_0}+w_{t_0}||}}$$
where
$$\displaystyle{v_t=((1-t)v_{0x}+tv_{1x},0)}$$
$$\displaystyle{w_t=(r_0(1-t)+tr_1)e^{i((1-t)\theta_0+t\theta_1)}}$$
with $\theta_1=\pi/2$ and $\theta_0=\pi/2+\arctan(5/\sqrt{299})$.
In fact, set $v_{0x}=4$, $v_{1x}=3.04$ $w_1=(0,2)$ and $w_0=\frac{2}{9}(-5,\sqrt{299})$.
Then $\frac{||v_1||}{||v_0||}=\frac 34+0.01=0.76$, $\frac{||w_1||}{||w_0||}=\frac 34$, and
$\frac{||v_1+w_1||}{||v_0+w_0||}=\frac{3\sqrt{13.2416}}{4\sqrt{13}}=0.756937...<0.76$.
But for $t_1=1/2$ and $t_0=0$ we have $||v_{t_1}||=3.52$, $||v_{t_0}||=4$, $||v_{t_0}+w_{t_0}||=\frac 43\sqrt{13}$,
$||v_{t_1}+w_{t_1}||=\frac{1}{2} \sqrt{85-84 \sin \left(\frac{1}{2} \tan ^{-1}\left(\frac{5}{\sqrt{299}}\right)\right)}=4.27834...$
and so
$\frac{||v_{t_1}||}{||v_{t_0}||}=0.88$ and $\frac{||v_1+w_1||}{||v_0+w_0||}=0.889949...$, so the inequality is not satisfied.
The strategy for finding a counterexample was to find $v_0,v_1,w_0,w_1$ such that
$$(1)\quad\quad\quad\frac{||v_1||}{||v_0||}= \frac{||w_1||}{||w_0||}\quad \text{and}\quad\frac{||v_1||}{||v_0||}= \frac{||v_1+w_1||}{||v_0+w_0||},$$
then for all $t\in[0,1]$ we should have the inequalites
$$
\frac{||w_1+v_1||}{||v_1||}\le \frac{||w_t+v_t||}{||v_t||}\le \frac{||w_0+v_0||}{||v_0||}=\frac{||w_1+v_1||}{||v_1||},
$$
but we find some $t$ that doesn't satisfies the inequalites, and then disturb the initial data a bit, so that we have inequalities in (1) but the same $t$ still contradicts the desired conclusion.
This strategy takes advantage of the fact that the deformation is not very well behaved.
