Start with the plane $\mathbb R^2$ and remove a slab, but keep a line going through the slab:

$$ Slab = \{(x, y) \in \mathbb R^2 : 0 < y < 1, x \neq 0\} $$ $$ M = \mathbb R^2 - Slab$$

  y

--------o-------------
--------o-------------
--------o-------------

x

Note that $M_1$ is connected but curves going from one side of the slab to the other have a fixed direction for some time.

Now cut away a line-with-a-hole:

$$ Line_\delta =\{(x, y) \in \mathbb R^2 : y = 1 + \delta, x \neq 50\}$$ $$ N_\delta = \mathbb R^2 - Slab - Line_\delta $$

  y
-----------------------------o--     -
                                     | delta
--------o-----------------------     -
--------o-----------------------
--------o-----------------------

x

At the top of the slab (the point $(1, 0)$) we will always have $\gamma'/|\gamma| = (0, 1)$. If $\delta$ is small enough compared to your $\epsilon$, you shouldn't be able to turn fast enough to avoid crashing into $Line_\delta$.

Edit: Another answer is to take the manifold

$$ ThickenedCircle_{r, \delta} = \{ p \in \mathbb R^2 : r-\delta < |p| < r+\delta \}.$$ First chose a sufficiently small $r$ so that the circle of radius $r$ does not obey your condition on the curve for $\epsilon/2$. Then if you chose $\delta$ small enough you get a flat incomplete 2-manifold where geodesics still must accelerate too much.

Start with the plane $\mathbb R^2$ and remove a slab, but keep a line going through the slab:

$$ Slab = \{(x, y) \in \mathbb R^2 : 0 < y < 1, x \neq 0\} $$ $$ M = \mathbb R^2 - Slab$$

  y

--------o-------------
--------o-------------
--------o-------------

x

Note that $M_1$ is connected but curves going from one side of the slab to the other have a fixed direction for some time.

Now cut away a line-with-a-hole:

$$ Line_\delta =\{(x, y) \in \mathbb R^2 : y = 1 + \delta, x \neq 50\}$$ $$ N_\delta = \mathbb R^2 - Slab - Line_\delta $$

  y
-----------------------------o--     -
                                     | δ
--------o-----------------------     -
--------o-----------------------
--------o-----------------------

x

At the top of the slab (the point $(1, 0)$) we will always have $\gamma'/|\gamma| = (0, 1)$. If $\delta$ is small enough compared to your $\epsilon$, you shouldn't be able to turn fast enough to avoid crashing into $Line_\delta$.

Edit: Another answer is to take the manifold

$$ ThickenedCircle_{r, \delta} = \{ p \in \mathbb R^2 : r-\delta < |p| < r+\delta \}.$$ First chose a sufficiently small $r$ so that the circle of radius $r$ does not obey your condition on the curve for $\epsilon/2$. Then if you chose $\delta$ small enough you get a flat incomplete 2-manifold where geodesics still must accelerate too much.

Start with the plane $\mathbb R^2$ and remove a slab, but keep a line going through the slab:

$$ Slab = \{(x, y) \in \mathbb R^2 : 0 < y < 1, x \neq 0\} $$ $$ M = \mathbb R^2 - Slab$$

  y

--------o-------------
--------o-------------
--------o-------------

x

Note that $M_1$ is connected but curves going from one side of the slab to the other have a fixed direction for some time.

Now cut away a line-with-a-hole:

$$ Line_\delta =\{(x, y) \in \mathbb R^2 : y = 1 + \delta, x \neq 50\}$$ $$ N_\delta = \mathbb R^2 - Slab - Line_\delta $$

  y
-----------------------------o--     -
                                     | delta
--------o-----------------------     -
--------o-----------------------
--------o-----------------------

x

At the top of the slab (the point $(1, 0)$) we will always have $\gamma'/|\gamma| = (0, 1)$. If $\delta$ is small enough compared to your $\epsilon$, you shouldn't be able to turn fast enough to avoid crashing into $Line_\delta$.

Start with the plane $\mathbb R^2$ and remove a slab, but keep a line going through the slab:

$$ Slab = \{(x, y) \in \mathbb R^2 : 0 < y < 1, x \neq 0\} $$ $$ M = \mathbb R^2 - Slab$$

  y

--------o-------------
--------o-------------
--------o-------------

x

Note that $M_1$ is connected but curves going from one side of the slab to the other have a fixed direction for some time.

Now cut away a line-with-a-hole:

$$ Line_\delta =\{(x, y) \in \mathbb R^2 : y = 1 + \delta, x \neq 50\}$$ $$ N_\delta = \mathbb R^2 - Slab - Line_\delta $$

  y
-----------------------------o--     -
                                     | delta
--------o-----------------------     -
--------o-----------------------
--------o-----------------------

x

At the top of the slab (the point $(1, 0)$) we will always have $\gamma'/|\gamma| = (0, 1)$. If $\delta$ is small enough compared to your $\epsilon$, you shouldn't be able to turn fast enough to avoid crashing into $Line_\delta$.

Edit: Another answer is to take the manifold

$$ ThickenedCircle_{r, \delta} = \{ p \in \mathbb R^2 : r-\delta < |p| < r+\delta \}.$$ First chose a sufficiently small $r$ so that the circle of radius $r$ does not obey your condition on the curve for $\epsilon/2$. Then if you chose $\delta$ small enough you get a flat incomplete 2-manifold where geodesics still must accelerate too much.