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As asked, the answer to the question is 'no'. The simply-connected cover $f:\mathbb{R}^2\to S$ of Sherck's first surface $S$ (which is defined in $\mathbb{R}^3$ by the equation $\mathrm{e}^{z} \cos x = \cos y$) has strictly negative curvature and is complete but there do not exist global coordinates $s,t$ on $\mathbb{R}^2$ (the domain of $f$) whose level sets are the asymptotic curves.

There are functions $u$ and $v$ on $S$ itself such that $(u,v):S\to\mathbb{R}^2$ is a coordinate chart on $S$ and the level curves of $u$ and of $v$ are (unions of) asymptotic curves on $S$. Explicitly, set $$ u(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{-}y)}{\cos x {+} \cos y}\right)\ \text{and}\ \ v(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{+}y)}{\cos x {+} \cos y}\right) $$ when $|x|,|y| <\pi/2$ and $(x,y,z)$ lies in $S$. These functions then extend analytically to the entire surface $S$ and have the claimed properties.

Now, $S$ is not simply-connected. The image $(u,v)(S)$ in $\mathbb{R}^2$ is the entire plane minus the points of the form $(2m{+}1,2n{+}1)$ where $m$ and $n$ are integers. Note, for example, that the lines $u=c$ and $v=c$ are asymptotic curves, but are disconnected if $c$ is an odd integer. (The case of $c$ an integer corresponds to vertical straight lines in $S$, i.e., where $\cos x = \cos y = 0$.)

It follows from this that, if $f:\mathbb{R}^2\to S$ is the simply-connected cover of $S$, then the functions $u{\circ}f$ and $v{\circ }f$ are not coordinates on $\mathbb{R}^2$, as they do not distinguish points in $\mathbb{R}^2$. If there were global asymptotic coordinates $s$ and $t$ on $\mathbb{R}^2$, then (after switching $s$ and $t$ if necessary) they would have to satisfy $\mathrm{d}s\wedge \mathrm{d}(u{\circ}f)=0$ and $\mathrm{d}t\wedge \mathrm{d}(v{\circ}f)=0$, and it is easy to see that no such functions can define a coordinate system and distinguish points.

In the general case, I think it will not be easy to formulate a necessary and sufficient condition for global asymptotic coordinates to exist on the simply-connected cover of an immersed surface of negative curvature in $\mathbb{R}^3$.

As asked, the answer to the question is 'no'. The simply-connected cover $f:\mathbb{R}^2\to S$ of Sherck's first surface $S$ (which is defined in $\mathbb{R}^3$ by the equation $\mathrm{e}^{z} \cos x = \cos y$) has strictly negative curvature and is complete but there do not exist global coordinates $s,t$ on $\mathbb{R}^2$ (the domain of $f$) whose level sets are the asymptotic curves.

There are functions $u$ and $v$ on $S$ itself such that $(u,v):S\to\mathbb{R}^2$ is a coordinate chart on $S$ and the level curves of $u$ and of $v$ are (unions of) asymptotic curves on $S$. Explicitly, set $$ u(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{-}y)}{\cos x {+} \cos y}\right)\ \text{and}\ \ v(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{+}y)}{\cos x {+} \cos y}\right) $$ when $|x|,|y| <\pi/2$ and $(x,y,z)$ lies in $S$. These functions then extend analytically to the entire surface $S$ and have the claimed properties.

It may help to observe that the surface $S$ is preserved by transformations $X$,$Y$,$D$ and $T$, where $X(x,y,z) = (-x,y,z)$, $Y(x,y,z)= (x,-y,z)$, $D(x,y,z)=(y,x,-z)$, and $T(x,y,z)=(x{+}\pi,y{+}\pi,z)$ and that $(u,v)\circ X = (-v,-u)$, $(u,v)\circ Y = (v,u)$, $(u,v)\circ D = (-u,v)$ and $(u,v)\circ T = (u,v{+}2)$ while $$ (u,v)(\pi/2,\pi/2,z) = \left(\frac2\pi\,\arcsin\bigl(\tanh(z/2)\bigr),1\right). $$ Application of these rules suffices to extend $u$ and $v$ to the entire surface $S$.

Now, $S$ is not simply-connected. The image $(u,v)(S)$ in $\mathbb{R}^2$ is the entire plane minus the points of the form $(2m{+}1,2n{+}1)$ where $m$ and $n$ are integers. Note, for example, that the lines $u=c$ and $v=c$ are asymptotic curves, but are disconnected if $c$ is an odd integer. (The case of $c$ an integer corresponds to vertical straight lines in $S$, i.e., where $\cos x = \cos y = 0$.)

It follows from this that, if $f:\mathbb{R}^2\to S$ is the simply-connected cover of $S$, then the functions $u{\circ}f$ and $v{\circ }f$ are not coordinates on $\mathbb{R}^2$, as they do not distinguish points in $\mathbb{R}^2$. If there were global asymptotic coordinates $s$ and $t$ on $\mathbb{R}^2$, then (after switching $s$ and $t$ if necessary) they would have to satisfy $\mathrm{d}s\wedge \mathrm{d}(u{\circ}f)=0$ and $\mathrm{d}t\wedge \mathrm{d}(v{\circ}f)=0$, and it is easy to see that no such functions can define a coordinate system and distinguish points.

In the general case, I think it will not be easy to formulate a necessary and sufficient condition for global asymptotic coordinates to exist on the simply-connected cover of an immersed surface of negative curvature in $\mathbb{R}^3$.

As asked, the answer to the question is 'no'. The simply-connected cover $f:\mathbb{R}^2\to S$ of Sherck's first surface $S$ (which is defined in $\mathbb{R}^3$ by the equation $\mathrm{e}^{z} \cos x = \cos y$) has strictly negative curvature and is complete but there do not exist global coordinates $s,t$ on $\mathbb{R}^2$ (the domain of $f$) whose level sets are the asymptotic curves.

There are functions $u$ and $v$ on $S$ itself such that $(u,v):S\to\mathbb{R}^2$ is a coordinate chart on $S$ and the level curves of $u$ and of $v$ are (unions of) asymptotic curves on $S$. Explicitly, set $$ u(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{-}y)}{\cos x {+} \cos y}\right)\ \text{and}\ \ v(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{+}y)}{\cos x {+} \cos y}\right) $$ when $|x|,|y| <\pi/2$ and $(x,y,z)$ lies in $S$. These functions then extend analytically to the entire surface $S$ and have the claimed properties.

Now, $S$ is not simply-connected. The image $(u,v)(S)$ in $\mathbb{R}^2$ is the entire plane minus the integer lattice $\mathbb{Z}^2$. Note, for example, that the lines $u=c$ and $v=c$ are asymptotic curves, but are disconnected if $c$ is an integer. (The case of $c$ an integer corresponds to vertical straight lines in $S$, i.e., where $\cos x = \cos y = 0$.)

It follows from this that, if $f:\mathbb{R}^2\to S$ is the simply-connected cover of $S$, then the functions $u{\circ}f$ and $v{\circ }f$ are not coordinates on $\mathbb{R}^2$, as they do not distinguish points in $\mathbb{R}^2$. If there were global asymptotic coordinates $s$ and $t$ on $\mathbb{R}^2$, then (after switching $s$ and $t$ if necessary) they would have to satisfy $\mathrm{d}s\wedge \mathrm{d}(u{\circ}f)=0$ and $\mathrm{d}t\wedge \mathrm{d}(v{\circ}f)=0$, and it is easy to see that no such functions can define a coordinate system and distinguish points.

In the general case, I think it will not be easy to formulate a necessary and sufficient condition for global asymptotic coordinates to exist on the simply-connected cover of an immersed surface of negative curvature in $\mathbb{R}^3$.

As asked, the answer to the question is 'no'. The simply-connected cover $f:\mathbb{R}^2\to S$ of Sherck's first surface $S$ (which is defined in $\mathbb{R}^3$ by the equation $\mathrm{e}^{z} \cos x = \cos y$) has strictly negative curvature and is complete but there do not exist global coordinates $s,t$ on $\mathbb{R}^2$ (the domain of $f$) whose level sets are the asymptotic curves.

There are functions $u$ and $v$ on $S$ itself such that $(u,v):S\to\mathbb{R}^2$ is a coordinate chart on $S$ and the level curves of $u$ and of $v$ are (unions of) asymptotic curves on $S$. Explicitly, set $$ u(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{-}y)}{\cos x {+} \cos y}\right)\ \text{and}\ \ v(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{+}y)}{\cos x {+} \cos y}\right) $$ when $|x|,|y| <\pi/2$ and $(x,y,z)$ lies in $S$. These functions then extend analytically to the entire surface $S$ and have the claimed properties.

Now, $S$ is not simply-connected. The image $(u,v)(S)$ in $\mathbb{R}^2$ is the entire plane minus the points of the form $(2m{+}1,2n{+}1)$ where $m$ and $n$ are integers. Note, for example, that the lines $u=c$ and $v=c$ are asymptotic curves, but are disconnected if $c$ is an odd integer. (The case of $c$ an integer corresponds to vertical straight lines in $S$, i.e., where $\cos x = \cos y = 0$.)

It follows from this that, if $f:\mathbb{R}^2\to S$ is the simply-connected cover of $S$, then the functions $u{\circ}f$ and $v{\circ }f$ are not coordinates on $\mathbb{R}^2$, as they do not distinguish points in $\mathbb{R}^2$. If there were global asymptotic coordinates $s$ and $t$ on $\mathbb{R}^2$, then (after switching $s$ and $t$ if necessary) they would have to satisfy $\mathrm{d}s\wedge \mathrm{d}(u{\circ}f)=0$ and $\mathrm{d}t\wedge \mathrm{d}(v{\circ}f)=0$, and it is easy to see that no such functions can define a coordinate system and distinguish points.

In the general case, I think it will not be easy to formulate a necessary and sufficient condition for global asymptotic coordinates to exist on the simply-connected cover of an immersed surface of negative curvature in $\mathbb{R}^3$.

As asked, the answer to the question is 'no'. The simply-connected cover $f:\mathbb{R}^2\to S$ of Sherck's first surface $S$ (which is defined in $\mathbb{R}^3$ by the equation $\mathrm{e}^{z} \cos y = \cos x$) has strictly negative curvature and is complete but there do not exist global coordinates $s,t$ on $\mathbb{R}^2$ (the domain of $f$) whose level sets are the asymptotic curves.

What's true is that there are functions $u$ and $v$ on $S$ itself such that $(u,v):S\to\mathbb{R}^2$ is a coordinate chart on $S$ and the level curves of $u$ and of $v$ are (unions of) asymptotic curves on $S$.

However, $S$ is not simply-connected; the image $(u,v)(S)$ in $\mathbb{R}^2$ is the entire plane minus the integer lattice $\mathbb{Z}^2$. It follows from this that, if $f:\mathbb{R}^2\to S$ is the simply-connected cover of $S$, then the functions $u{\circ}f$ and $v{\circ }f$ are not coordinates on $\mathbb{R}^2$, as they do not distinguish points. If there were global asymptotic coordinates $s$ and $t$ on $\mathbb{R}^2$, then (after switching $s$ and $t$ if necessary) they would have to satisfy $\mathrm{d}s\wedge \mathrm{d}(u{\circ}f)=0$ and $\mathrm{d}t\wedge \mathrm{d}(v{\circ}f)=0$, and it is easy to see that no such functions can define a coordinate system and distinguish points.

In the general case, I think it will not be easy to formulate a necessary and sufficient condition for global asymptotic coordinates to exist on the simply-connected cover of an immersed surface of negative curvature in $\mathbb{R}^3$.

As asked, the answer to the question is 'no'. The simply-connected cover $f:\mathbb{R}^2\to S$ of Sherck's first surface $S$ (which is defined in $\mathbb{R}^3$ by the equation $\mathrm{e}^{z} \cos x = \cos y$) has strictly negative curvature and is complete but there do not exist global coordinates $s,t$ on $\mathbb{R}^2$ (the domain of $f$) whose level sets are the asymptotic curves.

There are functions $u$ and $v$ on $S$ itself such that $(u,v):S\to\mathbb{R}^2$ is a coordinate chart on $S$ and the level curves of $u$ and of $v$ are (unions of) asymptotic curves on $S$. Explicitly, set $$ u(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{-}y)}{\cos x {+} \cos y}\right)\ \text{and}\ \ v(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{+}y)}{\cos x {+} \cos y}\right) $$ when $|x|,|y| <\pi/2$ and $(x,y,z)$ lies in $S$. These functions then extend analytically to the entire surface $S$ and have the claimed properties.

Now, $S$ is not simply-connected. The image $(u,v)(S)$ in $\mathbb{R}^2$ is the entire plane minus the integer lattice $\mathbb{Z}^2$. Note, for example, that the lines $u=c$ and $v=c$ are asymptotic curves, but are disconnected if $c$ is an integer. (The case of $c$ an integer corresponds to vertical straight lines in $S$, i.e., where $\cos x = \cos y = 0$.)

It follows from this that, if $f:\mathbb{R}^2\to S$ is the simply-connected cover of $S$, then the functions $u{\circ}f$ and $v{\circ }f$ are not coordinates on $\mathbb{R}^2$, as they do not distinguish points in $\mathbb{R}^2$. If there were global asymptotic coordinates $s$ and $t$ on $\mathbb{R}^2$, then (after switching $s$ and $t$ if necessary) they would have to satisfy $\mathrm{d}s\wedge \mathrm{d}(u{\circ}f)=0$ and $\mathrm{d}t\wedge \mathrm{d}(v{\circ}f)=0$, and it is easy to see that no such functions can define a coordinate system and distinguish points.

In the general case, I think it will not be easy to formulate a necessary and sufficient condition for global asymptotic coordinates to exist on the simply-connected cover of an immersed surface of negative curvature in $\mathbb{R}^3$.