Forget to begin with that we have to find a path, but simply try covering the $9$ points by four infinte straight lines.
Call a line covering $n$ points an $n$-line. Since four lines covering only $2$ points each can never cover $9$ points a solution has at least one $3$-line in it. When identifying rotations and reflections we have exactly three distinct types of $3$-lines:
No parallel $3$-lines
It is not possible to have a solution involving parallel $3$-lines. For instance if we have the following situation:
When the two blue $3$-lines shown above have been placed we have to cover the remaining three points by two lines. So at least one line must cover more than one point which implies that the dashed grey line has to be added. But three parallel lines are clearly impossible, since it takes two additional linear moves to visit and travel along all three, but we only have one line left!
Ruling out type 3
With the above we can rule out type 3 being a part of any solution: We cannot have any line parallel to the initial $3$-line by the above argument. This actually means that each of the remaining three lines has to cover exactly $2$ of the remaining $6$ points each. Also we can only have one line orthogonal to the $3$-line for otherwise we would again have parallel $3$-lines (in the orthogonal direction).
So we have to have two lines passing through $2$ of the remaining $6$ points each that are not parallel nor orthogonal to the initial $3$-line of type 3. When symmetries are considered, this gives essentially two setups to consider. The first one is:
which clearly does not hold a solution since the lines only intersect at one point so it takes far more than four linear moves to travel since you would have to return to this intersection time and again in any path traversing parts of all of these lines. The other setup to consider is:
One way of seing why the above holds no solution is, that some of the lines are only intersected in ways having some of their points lying to both sides of the point of intersection. This makes it impossible to go through all points on any of these without visiting that line twice or reversing direction at some point.
Ruling out exclusive type 1 or 2
If we start from type 1 and do not allow type 2 and already know that type 3 must be ruled out and that no line parallel to the initial $3$-line is allowed, we see that this means that a line that passes the center point of the $9$ point pattern can pass no other point it the pattern as that would lead to one of the situations just mentioned. But with a line that only passes through the center the remaining two lines must covers $5$ points, so one of the lines must cover the $3$ points that are parallel to the initial $3$-line which again leads to a paralel $3$-lines contradiction.
If we start from type 2 and do not allow the other two types, there is essentially only one setup not already covered to consider:
which is another setup having three parallel lines. By previous arguments three distinct parallel lines can never be part of any solution.
The unique solution (type 1 + type 2)
By inclusion-exclusion any solution must involve type 1 and type 2 at the same time. So the situation looks like this to begin with:
and since we already know that we are not allowed to add a line parallel to the $3$-line of type 1 there are only two ways to complete the above setup. One way is like this:
and we see how the $3$-line of type 1 is the only line connecting to the other three lines to one side of their points. The other three lines divide each other in parts having points on each side. So all three other lines must be connected to the $3$-line of type 1 in a given solution. This is impossible since each line can only be connected to two other lines in a path that does not travel along any line twice.
So finally this leaves us with the well known solution as essentially the only possibility when considering rotations and reflections.
I found this not hard to solve with a pen and a piece of paper towel, so I suspect that it's rather unconstrained and there are many solutions. Later on I may write a program to generate all possible solutions and count them, but for now here's what I did to find a single solution. (Addendum: I did eventually write the program; see the comment below.)
The first step, as already described in the comments, is to find out what the sum of the numbers on a face must be. Let $F$ be the sum of the four edges of a face. The total of all six faces is $6F$. But this counts each edge twice, and the sum of the 12 edges is $1+2+\cdots+12 = 78$, so $6F = 2\cdot 78$, and $$F=26.$$
(Consider a related problem: Arrange the numbers $1,\ldots,12$ in a $3\times 4$ array so that the numbers in each row have the same sum and the numbers in each column have the same sum. There are 4 columns, whose sums together add up to $1+2+\cdots+12 =78$, so for each to have the same sum, each one must sum to $\frac{78}4$. This is impossible, so we know right away that there is no solution.)
With $F=26$, it's natural to consider the numbers in pairs $1–12, 2–11, \ldots, 6–7$ where the pairs always add up to $13$. If we could arrange that each face had exactly two pairs, we would win. Also, we might imagine that the numbers are divided into small numbers $1\ldots 6$ and large numbers $7\ldots 12$. Maybe we can find a solution by matching small numbers with large numbers. To that end, let's arrange $1,12,2,11$ around the middle of the cube, like this:
This drawing is too complicated and not compact enough. If we could find a compact representation of the cube that still shows the relationships between the edges and the faces. it would be quicker and simpler to investigate different arrangements of numbers. So suppose the edges of the cube are labeled like this:
From now on we'll write this cube like this:
$$\def\r#1{\color{maroon}{#1}}\begin{array}{cccccccc}
& \r E && \r F && \r G && \r H\\\r A && \r B && \r C && \r D \\& \r I && \r J && \r K && \r L \\
\end{array}$$
Five of the six faces are easily visible in this display. Three of the middle faces are $AEBI, BFCJ, CGDK$. The fourth wraps around from the right end to the left: $DHAL$. The last two faces are the top and bottom and are $EFGH$ and $IJKL$, which are easy to see in the diagram.
Our assignment of $1,12,2,11$ from before is now:
$$\def\bl#1{\color{blue}{#1}}
\begin{array}{cccccccc}
& \r E && \r F && \r G && \r H\\\bl1 && \bl{12} && \bl2 && \bl{11} \\& \r I && \r J && \r K && \r L \\
\end{array}$$
The four side faces have so far been assigned edges that total 13, 14, 13, and 12 respectively. We have remaining the four small numbers $3,4,5,6$ and the four large numbers $7,8,9,10$. Since $E$ and $I$ must total 13, let's assign them $6$ and $7$ and see what happens next:
$$\begin{array}{cccccccc}
& \bl6 && \r F && \r G && \r H\\1 && 12 && 2 && 11 \\& \bl7 && \r J && \r K && \r L \\
\end{array}$$
Now we need to assign $F$ and $J$, which must add to 12. We have left $3,4,5,8,9,10$. We could use $3,9$ or $4,8$. Let's try $3,9$. (Let's call this decision “$\star$”; it turns out to be wrong, so we'll come back to it later.) It doesn't really matter which one we put on top, since we can switch them later without affecting any of the totals except the top and bottom face. But since we put the small number $6$ on the top face before, let's put the large number $9$ on top this time, to try to keep the top and bottom faces balanced:
$$\begin{array}{cccccccc}
& 6 && \bl9 && \r G && \r H\\1 && 12 && 2 && 11 \\& 7 && \bl3 && \r K && \r L \\
\end{array}$$
Now we want to assign $G$ and $K$. We have $G+2+K+11 = 26$, so $G+K= 13$, and the unused numbers are $4,5,8,10$. The only pair we can still use is $5,8$. So let's assign those:
$$\begin{array}{cccccccc}
& 6 && 9 && \bl5 && \r H\\1 && 12 && 2 && 11 \\& 7 && 3 && \bl8 && \r L \\
\end{array}$$
Finally, the last two numbers are $4$ and $10$, which combine with the $11$ and $1$ in the $11,L,1,H$ face to make 26. Since we put the small number $5$ on top last time, let's put the large number $10$ on top this time:
$$\begin{array}{cccccccc}
& 6 && 9 && 5 && \bl{10}\\1 && 12 && 2 && 11 \\& 7 && 3 && 8 && \bl4 \\
\end{array}$$
Now the four faces around the middle all add up to 26, but the top face is $6+9+5+10 = 30$ and the bottom face is $7+3+8+4 = 22$. Let's call this a “deficit” of 8, the amount by which the bottom face falls short of the amount of the top face. We want a deficit of 0.
We can change the deficit by swapping numbers between the top and bottom. The sides faces all add to 26 so we don't want to change them. But if we swap a top number $t$ with the corresponding bottom number $b$ on the same side face, the deficit will decrease by $t-b$, and the totals on all four side faces will remain as they were. The top-bottom pairs are $6–7, 9–3, 5–8, $ and $10–4$. These contribute $-1, +6, -3, $ and $+6$ to the deficit, respectively, for a total of $+8$. We can change the sign of each contribution from $+$ to $-$ or vice versa, and we want the total deficit to be 0. This is evidently impossible (to see this, just consider whether the 6es should have the same sign or opposite signs; neither works), so we are in a blind alley.
Backing up to the last choice we had, $\star$, we found that adding $3,9$ at that point failed. So let's add $4,8$ instead, obtaining:
$$\begin{array}{cccccccc}
& 6 && \bl8 && \r G && \r H\\1 && 12 && 2 && 11 \\& 7 && \bl4 && \r K && \r L \\
\end{array}$$
Then $G+K= 13$ and we have only $3 ,5, 9, 10$, so we take $G,K =3,10$:
$$\begin{array}{cccccccc}
& 6 && 8 && \bl3 && \r H\\1 && 12 && 2 && 11 \\& 7 && 4 && \bl{10} && \r L \\
\end{array}$$
and then $5,9$ go in last:
$$\begin{array}{cccccccc}
& 6 && 8 && 3 && \bl5\\1 && 12 && 2 && 11 \\& 7 && 4 && 10 && \bl9 \\
\end{array}$$
This time the top totals $6+8+3+5= 22$ and the bottom $7+4+10+9 = 30$, so the bottom has a surplus of $8$. The top-bottom pairs contribute surpluses of $+1, -4, +7, +4$, respectively. We want to switch the signs on these so that they add up to 0. Clearly switching the $+4$ to $-4$ will work, so we swap the positions of $5$ and $9$:
$$\begin{array}{cccccccc}
& 6 && 8 && 3 && \bl9\\1 && 12 && 2 && 11 \\& 7 && 4 && 10 && \bl5 \\
\end{array}$$
which is a solution:
Best Answer
To solve it by hand you need to look for words that have similar sets of letters. Using OREGON and RENO you know $G+O=28$. It's too bad they didn't give you ARKANSAS. RENO and NOME give $M=R+3$. Can you find $D+A-O=17?$ MONTGOMERY and MONTEREY are interesting. It is supposed to be a certain amount of work.