I think the rule is

• The car has the initial velocity of (0, 0).
• At every step, you can accelerate by +1, 0, or -1 unit, independently for X and Y directions, and then change position according to the new velocity.

The paths for levels 1 to 11 are the same as in Jujustum's answer, but some require an alternative explanation.

To ease explanation, let's define

right is the positive X direction, and down is the positive Y direction.

Level 6

Accelerate +1 in both X and Y directions, twice.

Level 7

First, accelerate +1 in both X and Y directions. Then accelerate +1 in X and -1 in Y to get the velocity of (2, 0) and reach the goal.

Level 8

Same as Level 7 for the first two steps. Then, accelerate -1 in X and -1 in Y to get the velocity of (1, -1).

Level 10

Accelerate -1 in X to get the velocity of (-1, 0) and go back 1 square. Then accelerating +1 in X gives the velocity of (0, 0), keeping the car in place. Now accelerate +1 twice to reach the goal.

Level 13

I can't upload an image at the moment, so I'll share a penpa link instead. This path takes 7 steps.

• The square labeled 1 (the one right below the start) cannot be avoided since the first acceleration in Y will put you somewhere on that row.
• Then the path tries to accelerate in X as much as possible. I think velocity of 3 in X is not reachable, so the path goes on with the constant X velocity of 2. Y velocity is then adjusted to keep the car on the land and reach the goal with the Y velocity of 2.

Level 14

Again, here is a penpa link in place of an image. This path takes 8 steps.

• This time, the focus is on maximizing the X acceleration. Accleration of +1 twice in X from the start is impossible, so the car accelerates +1 and then +0, and then keeps accelerating until it builds up X velocity up to 5.
• On the process, the car happens to nicely jump onto the second line at step 4, and step on the middle of I at velocity 5.
• The optimality of this path can be proven by $1 + 1 + 2 + 3 + 4 + 5 + 6 = 22 < 25$, which is the theoretical maximum X displacement in 7 steps under the constraint that accelerating twice in 2 steps is impossible.

I think the rule is

• The car has the initial velocity of (0, 0).
• At every step, you can accelerate by +1, 0, or -1 unit, independently for X and Y directions, and then change position according to the new velocity.

The paths for levels 1 to 11 are the same as in Jujustum's answer, but some require an alternative explanation.

To ease explanation, let's define

right is the positive X direction, and down is the positive Y direction.

Level 6

Accelerate +1 in both X and Y directions, twice.

Level 7

First, accelerate +1 in both X and Y directions. Then accelerate +1 in X and -1 in Y to get the velocity of (2, 0) and reach the goal.

Level 8

Same as Level 7 for the first two steps. Then, accelerate -1 in X and -1 in Y to get the velocity of (1, -1).

Level 10

Accelerate -1 in X to get the velocity of (-1, 0) and go back 1 square. Then accelerating +1 in X gives the velocity of (0, 0), keeping the car in place. Now accelerate +1 twice to reach the goal.

Level 13

I can't upload an image at the moment, so I'll share a penpa link instead. This path takes 7 steps.

• The square labeled 1 (the one right below the start) cannot be avoided since the first acceleration in Y will put you somewhere on that row.
• Then the path tries to accelerate in X as much as possible. I think velocity of 3 in X is not reachable, so the path goes on with the constant X velocity of 2. Y velocity is then adjusted to keep the car on the land and reach the goal with the Y velocity of 2.

Level 14

Again, here is a penpa link in place of an image. This path takes 8 steps.

• This time, the focus is on maximizing the X acceleration. Accleration of +1 twice in X from the start is impossible, so the car accelerates +1 and then +0, and then keeps accelerating until it builds up X velocity up to 5.
• On the process, the car happens to nicely jump onto the second line at step 4, and step on the middle of I at velocity 5.
• The optimality of this path can be proven by $1 + 1 + 2 + 3 + 4 + 5 + 6 = 22 < 25$, which is the theoretical maximum X displacement in 7 steps under the constraint that accelerating twice in the first two steps is impossible.