The answer is yes, here is a proof.

Let $k=2n$. The polynomial, regarded as a map $f:\mathbb R^k\to\mathbb R^k$, has the following properties:

(1) The pre-image of every point if finite, moreover the cardinality of a pre-image is uniformly bounded by some constant $N$ (by Bezout's theorem, see comments).

(2) The set $\Sigma$ of singularities of $f$ (i.e. points where $\det df=0$) is a union of finitely many smooth submanifolds of (real) codimension at least 2.

This (along with smoothness and properness) implies the path-lifting property.

First, observe the following facts:

(3) $f(\Sigma)$ is closed (because $\Sigma$ is closed and $f$ is proper).

(4) $\mathbb R^k\setminus f(\Sigma)$ is path connected. Moreover any path in $\mathbb R^k$ is a uniform limit of paths avoiding $f(\Sigma)$. This follows from the fact that $f(\Sigma)$ has Hausdorff dimension at most $k-2$.

(5) For every compact set $K\subset\mathbb R^k$ and every $\varepsilon>0$, there exists $\delta>0$ such that for every connected set $S\subset K$ of diameter greater than $\delta$, the diameter of $f(S)$ is greater than $\varepsilon$. Indeed, if this is not the case, there would be a sequence $S_i$ of connected subsets of $K$ with diameters at least some $\varepsilon_0>0$ and diameters of images going to zero. By choosing a subsequence, we may assume that the images converge to some $y_0\in\mathbb R^k$. Since $S_i$ is connected and has diameter at least $\varepsilon_0$, it contains a finite subset $P_i$ of cardinality $N+1$ such that all points of $P_i$ are separated away from one another by distance at least $\varepsilon_0/2N$. A subsequence of $\{P_i\}$ converge to a set $P$ of cardinality $N+1$, and all points of $P$ are mapped by $f$ to $x_0$, contrary to (1).

Now, in order to lift a path $\gamma$, approximate it by paths $\gamma_i\subset\mathbb R^k\setminus f(\Sigma)$. The restriction of $f$ to the pre-image of $R^k\setminus f(\Sigma)$ is a covering map to $R^k\setminus f(\Sigma)$ because it is a proper local homeomorphism. So there are lifts $\tilde\gamma_i$ of $\gamma_i$. Now it suffices to find a converging subsequence of $\{\tilde\gamma_i\}$. To do this, it suffices to show that this sequence is equi-continuous. And this follows from (5): if you could find arbitrarily close pair of points on some $\gamma_i$ with lifts bounded away from each other, the lift of the segment of $\gamma_i$ between these points would have diameter bounded away from zero but the diameter of the image arbitrarily small, contrary to (5).

To make sure that the limit lift starts from $z_0$, choose $\gamma_i$'s so that their starting points have pre-images near $z_0$ and start $\tilde\gamma_i$'s from there.

The answer is yes, here is a proof.

Let $k=2n$. The polynomial, regarded as a map $f:\mathbb R^k\to\mathbb R^k$, has the following properties:

(1) The pre-image of every point if finite, moreover the cardinality of a pre-image is uniformly bounded by some constant $N$ (by Bezout's theorem, see comments).

(2) The set $\Sigma$ of singularities of $f$ (i.e. points where $\det df=0$) is a union of finitely many smooth submanifolds of (real) codimension at least 2.

This (along with smoothness and properness) implies the path-lifting property.

First, observe the following facts:

(3) $f(\Sigma)$ is closed (because $\Sigma$ is closed and $f$ is proper).

(4) $\mathbb R^k\setminus f(\Sigma)$ is path connected. Moreover any path in $\mathbb R^k$ is a uniform limit of paths avoiding $f(\Sigma)$. This follows from the fact that $f(\Sigma)$ has Hausdorff dimension at most $k-2$.

(5) For every compact set $K\subset\mathbb R^k$ and every $\varepsilon>0$, there exists $\delta>0$ such that for every connected set $S\subset K$ of diameter greater than $\varepsilon$, the diameter of $f(S)$ is greater than $\delta$. Indeed, if this is not the case, there would be a sequence $S_i$ of connected subsets of $K$ with diameters at least some $\varepsilon_0>0$ and diameters of images going to zero. By choosing a subsequence, we may assume that the images converge to some $y_0\in\mathbb R^k$. Since $S_i$ is connected and has diameter at least $\varepsilon_0$, it contains a finite subset $P_i$ of cardinality $N+1$ such that all points of $P_i$ are separated away from one another by distance at least $\varepsilon_0/2N$. A subsequence of $\{P_i\}$ converge to a set $P$ of cardinality $N+1$, and all points of $P$ are mapped by $f$ to $x_0$, contrary to (1).

Now, in order to lift a path $\gamma$, approximate it by paths $\gamma_i\subset\mathbb R^k\setminus f(\Sigma)$. The restriction of $f$ to the pre-image of $R^k\setminus f(\Sigma)$ is a covering map to $R^k\setminus f(\Sigma)$ because it is a proper local homeomorphism. So there are lifts $\tilde\gamma_i$ of $\gamma_i$. Now it suffices to find a converging subsequence of $\{\tilde\gamma_i\}$. To do this, it suffices to show that this sequence is equi-continuous. And this follows from (5): if you could find arbitrarily close pair of points on some $\gamma_i$ with lifts bounded away from each other, the lift of the segment of $\gamma_i$ between these points would have diameter bounded away from zero but the diameter of the image arbitrarily small, contrary to (5).

To make sure that the limit lift starts from $z_0$, choose $\gamma_i$'s so that their starting points have pre-images near $z_0$ and start $\tilde\gamma_i$'s from there.

The answer is yes, here is a proof.

Let $k=2n$. The polynomial, regarded as a map $f:\mathbb R^k\to\mathbb R^k$, has the following properties:

(1) The pre-image of every point if finite, moreover the cardinality of a pre-image is uniformly bounded by some constant $N$ (by Bezout's theorem).

(2) The set $\Sigma$ of singularities of $f$ (i.e. points where $\det df=0$) is a union of finitely many smooth submanifolds of (real) codimension at least 2.

This (along with smoothness and properness) implies the path-lifting property.

First, observe the following facts:

(3) $f(\Sigma)$ is closed (because $\Sigma$ is closed and $f$ is proper).

(4) $\mathbb R^k\setminus f(\Sigma)$ is path connected. Moreover any path in $\mathbb R^k$ is a uniform limit of paths avoiding $f(\Sigma)$. This follows from the fact that $f(\Sigma)$ has Hausdorff dimension at most $k-2$.

(5) For every compact set $K\subset\mathbb R^k$ and every $\varepsilon>0$, there exists $\delta>0$ such that for every connected set $S\subset K$ of diameter greater than $\delta$, the diameter of $f(S)$ is greater than $\varepsilon$. Indeed, if this is not the case, there would be a sequence $S_i$ of connected subsets of $K$ with diameters at least some $\delta_0>0$ and diameters of images going to zero. By choosing a subsequence, we may assume that the images converge to some $y_0\in\mathbb R^k$. Since $S_i$ is connected and has diameter at least $\delta_0$, it contains a finite subset $P_i$ of cardinality $N+1$ such that all points of $P_i$ are separated away from one another by distance at least $\delta_0/2N$. A subsequence of $\{P_i\}$ converge to a set $P$ of cardinality $N+1$, and all points of $P$ are mapped by $f$ to $x_0$, contrary to (1).

Now, in order to lift a path $\gamma$, approximate it by paths $\gamma_i\subset\mathbb R^k\setminus f(\Sigma)$. The restriction of $f$ to the pre-image of $R^k\setminus f(\Sigma)$ is a covering map to $R^k\setminus f(\Sigma)$ because it is a proper local homeomorphism. So there are lifts $\tilde\gamma_i$ of $\gamma_i$. Now it suffices to find a converging subsequence of $\{\tilde\gamma_i\}$. To do this, it suffices to show that this sequence is equi-continuous. And this follows from (5): if you could find arbitrarily close pair of points on some $\gamma_i$ with lifts bounded away from each other, the lift of the segment of $\gamma_i$ between these points would have diameter bounded away from zero but the diameter of the image arbitrarily small, contrary to (5).

To make sure that the limit lift starts from $z_0$, choose $\gamma_i$'s so that their starting points have pre-images near $z_0$ and start $\tilde\gamma_i$'s from there.

The answer is yes, here is a proof.

Let $k=2n$. The polynomial, regarded as a map $f:\mathbb R^k\to\mathbb R^k$, has the following properties:

(1) The pre-image of every point if finite, moreover the cardinality of a pre-image is uniformly bounded by some constant $N$ (by Bezout's theorem, see comments).

(2) The set $\Sigma$ of singularities of $f$ (i.e. points where $\det df=0$) is a union of finitely many smooth submanifolds of (real) codimension at least 2.

This (along with smoothness and properness) implies the path-lifting property.

First, observe the following facts:

(3) $f(\Sigma)$ is closed (because $\Sigma$ is closed and $f$ is proper).

(4) $\mathbb R^k\setminus f(\Sigma)$ is path connected. Moreover any path in $\mathbb R^k$ is a uniform limit of paths avoiding $f(\Sigma)$. This follows from the fact that $f(\Sigma)$ has Hausdorff dimension at most $k-2$.

(5) For every compact set $K\subset\mathbb R^k$ and every $\varepsilon>0$, there exists $\delta>0$ such that for every connected set $S\subset K$ of diameter greater than $\delta$, the diameter of $f(S)$ is greater than $\varepsilon$. Indeed, if this is not the case, there would be a sequence $S_i$ of connected subsets of $K$ with diameters at least some $\varepsilon_0>0$ and diameters of images going to zero. By choosing a subsequence, we may assume that the images converge to some $y_0\in\mathbb R^k$. Since $S_i$ is connected and has diameter at least $\varepsilon_0$, it contains a finite subset $P_i$ of cardinality $N+1$ such that all points of $P_i$ are separated away from one another by distance at least $\varepsilon_0/2N$. A subsequence of $\{P_i\}$ converge to a set $P$ of cardinality $N+1$, and all points of $P$ are mapped by $f$ to $x_0$, contrary to (1).

Now, in order to lift a path $\gamma$, approximate it by paths $\gamma_i\subset\mathbb R^k\setminus f(\Sigma)$. The restriction of $f$ to the pre-image of $R^k\setminus f(\Sigma)$ is a covering map to $R^k\setminus f(\Sigma)$ because it is a proper local homeomorphism. So there are lifts $\tilde\gamma_i$ of $\gamma_i$. Now it suffices to find a converging subsequence of $\{\tilde\gamma_i\}$. To do this, it suffices to show that this sequence is equi-continuous. And this follows from (5): if you could find arbitrarily close pair of points on some $\gamma_i$ with lifts bounded away from each other, the lift of the segment of $\gamma_i$ between these points would have diameter bounded away from zero but the diameter of the image arbitrarily small, contrary to (5).

To make sure that the limit lift starts from $z_0$, choose $\gamma_i$'s so that their starting points have pre-images near $z_0$ and start $\tilde\gamma_i$'s from there.