**Edit:** In response to asker comment, edited significatively the answer as it failed to take into account the case $n<k$

Here is my argument why the probability of linear independence is 1:

Case $k\leq n$:

Notice $y_1,...,y_k$ linearly dependent implies $x_1,...,x_k$ are linearly dependent. So the probability that $y_1,...,y_k$ are linearly dependent is less or equal than the probability that $x_1,...,x_k$ are linearly dependent. In this case, the probability of $x_1,...,x_k$ being linearly dependent is 0.

The proof is similar to one answer to this question. The idea is that, for each $l\leq k$, the probability that $x_l$ is in the same subspace than $x_1,...,x_{l-1}$ is 0.

Case $n<k\leq 2n$:

Let $x_{1},...,x_{k}$ s.t. $y_{1},...,y_{k}$ are linearly dependent. Since the probability that $x_{1},...,x_{n}$ are linearly independent is 1, we can assume w.l.o.g $x_{1},...,x_{n}$ will be linearly independent. Then $x_{1},...,x_{n}$ are a basis of $\mathbb{R}^{n}$. In particular, for $l\in\{n+1,...,k\}$, there are unique $w_{1l},...,w_{nl}$ s.t.
$x_{l}=\sum_{i=1}^{n}w_{il}x_{i}$.

Now, since $y_{1},...,y_{k}$ are linearly dependent, there is $
\lambda\in\mathbb{C}^{k}\setminus\{0\}$ s.t.
$$
0=\sum_{i=1}^{k}\lambda_{i}y_{i}=\sum_{i=1}^{k}\lambda_{i}\left(\begin{array}{c}
x_{i}\\
\varphi_{k}^{i}x_{i}
\end{array}\right)
$$

Expanding $$
0=\sum_{i=1}^{k}\lambda_{i}\left(\begin{array}{c}
x_{i}\\
\varphi_{k}^{i}x_{i}
\end{array}\right)
$$
we get
\begin{eqnarray}
0 & = & \sum_{i=1}^{n}\lambda_{i}\left(\begin{array}{c}
x_{i}\\
\varphi_{k}^{i}x_{i}
\end{array}\right)+\sum_{l=n+1}^{k}\lambda_{l}\left(\begin{array}{c}
x_{l}\\
\varphi_{k}^{l}x_{l}
\end{array}\right)
\\
& = & \sum_{i=1}^{n}\lambda_{i}\left(\begin{array}{c}
x_{i}\\
\varphi_{k}^{i}x_{i}
\end{array}\right)+\sum_{l=n+1}^{k}\lambda_{l}\left(\begin{array}{c}
\sum_{i=1}^n w_{il}x_i\\
\varphi_{k}^{l}\sum_{i=1}^n w_{il}x_i
\end{array}\right)\\
& = & \sum_{i=1}^{n}\left(\begin{array}{c}
x_{i}\left(\lambda_i+\sum_{l=n+1}^k\lambda_lw_{il}\right)\\
x_{i}\left(\varphi_{k}^i\lambda_i+\sum_{l=n+1}^k\lambda_l\varphi_k^lw_{il}\right)
\end{array}\right)
\end{eqnarray}
Using $x_1,...,x_n$ are linearly independent, we deduce, for all $i\in\{1,...,n\}$,
\begin{eqnarray}
\lambda_{i}+\sum_{l=n+1}^{k}\lambda_{l}w_{il} & = & 0\\
\lambda_{i}+\sum_{l=n+1}^{k}\lambda_{l}\varphi_{k}^{l-i}w_{il} & = & 0
\end{eqnarray}
which implies $\sum_{l=n+1}^{k}w_{il}\lambda_{l}(1-\varphi_{k}^{l-i})=0$
and $(\lambda_{n+1},...,\lambda_{k})\neq0$. In other words, for all $i$, $(w_{in+1},...,w_{ik})$ are linearly dependent, which happens with probability 0.