Deducing the gradient of the functional $L(\gamma)=\int_0^1\|\gamma'(s)\|\ \mathrm ds$ is of the form $\nabla L(s)=-\kappa(s)\nu(s)$ We have $ \gamma : [ 0 , 1 ] \to \mathbb R ^ 2 $ as a parametrized curve such that its first and second order derivatives $ \gamma ' ( s ) $, $ \gamma '' ( s ) $ are defined and continuous, and such that $ \gamma ' ( s ) $ is different from $ 0 $ for all $ s \in [ 0 , 1 ] $ (a regular curve).
We assume that $ \gamma $ is a closed curve: $ \gamma $ and its derivatives take the same values at $ 0 $ and $ 1 $.
We have shown that for every $ s \in [ 0 , 1 ] $, $ \tau ' ( s ) $ (the derivative of $ \tau ( s ) $) is collinear to $ \nu ( s ) $, so there exists $ \kappa ( s ) \in \mathbb R $ such that $ \tau ' ( s ) = \kappa ( s ) \nu ( s ) $. $ \tau $ is the unit tangent vector ($ \tau ( s ) = \frac { \gamma ' ( s ) } { \| \gamma ' ( s ) \| } $) and $ \nu ( s ) $ is the unit normal vector (the vector obtained by applying a $ \frac \pi 2 $ rotation to $ \tau ( s ) $)
$ \kappa ( s ) $ is called the curvature of the curve $ \gamma $ at location $ \gamma ( s ) $.
We consider the following functional: $ L ( \gamma ) = \int _ 0 ^ 1 \| \gamma ' ( s ) \| \ \mathrm d s $.
We have considered a variation of the curve $ \gamma $, indexed by $ t \in \mathbb R $: $ \gamma _ t ( s ) = \gamma ( s ) + t u ( s ) $, where $ u ( s ) $ is a regular function from $ [ 0 , 1 ] $ to $ \mathbb R ^ 2 $. We have calculated the derivative of $ L ( \gamma _ t ) $ at $ t = 0 $, and shown that it is a linear function of $ u $, so $ \frac { \mathrm d } { \mathrm d t } L ( \gamma _ t ) = \psi ( u ) $, $ \psi $ being linear. $ \psi $ corresponds to the differential of $ L $ at $ \gamma $: $ \psi = \mathrm d L ( \gamma ) $.
Using integration by parts, we would like to deduce that the gradient of $ L $, i.e. the function $ \nabla L : [ 0 , 1] \to \mathbb R ^ 2 $ such that $ \psi ( u ) = \int _ 0 ^ 1 \langle \nabla L , u ( s ) \rangle $, is given by:
$$ \nabla L ( s ) = - \kappa ( s ) \nu ( s ) $$
Can someone please help?
 A: It seems that the task is very easy, and all the complications you're facing come from confusing some definitions and abusing some notations. I'll try to list some of them. But first, Let's make the simple calculations (including the integration by parts mentioned by yourself) needed for finding what you seek:
$$ \psi ( u ) = \left. \frac { \mathrm d } { \mathrm d t } \int _ 0 ^ 1 \| \gamma ' ( s ) + t u ' ( s )  \| \ \mathrm d s \right| _ { t = 0 } \\
= \left. \int _ 0 ^ 1 \frac { \partial } { \partial t } \left( \sqrt { \big( \gamma ' ( s ) + t u ' ( s ) \big) \cdot \big( \gamma ' ( s ) + t u ' ( s ) \big) } \right) \ \mathrm d s \right| _ { t = 0 } \\
= \left. \int _ 0 ^ 1 \frac { \big( \gamma ' ( s ) + t u ' ( s ) \big) \cdot \frac { \partial } { \partial t } \big( \gamma ' ( s ) + t u ' ( s ) \big) } { \sqrt { \big( \gamma ' ( s ) + t u ' ( s ) \big) \cdot \big( \gamma ' ( s ) + t u ' ( s ) \big) } } \ \mathrm d s \right| _ { t = 0 } \\
= \left. \int _ 0 ^ 1 \frac { \gamma ' ( s ) + t u ' ( s ) } { \| \gamma ' ( s ) + t u ' ( s ) \| } \cdot u ' ( s ) \ \mathrm d s \right| _ { t = 0 } \\
= \int _ 0 ^ 1 \frac { \gamma ' ( s ) } { \| \gamma ' ( s ) \| } \cdot u ' ( s ) \ \mathrm d s = \int _ 0 ^ 1 \tau ( s ) \cdot u ' ( s ) \ \mathrm d s \\
= \left[ \tau ( s ) \cdot u ( s ) \right] _ { s = 0 } ^ 1 - \int _ 0 ^ 1 \tau ' ( s ) \cdot u ( s ) \ \mathrm d s \\
= \left[ \tau ( s ) \cdot u ( s ) \right] _ { s = 0 } ^ 1 - \int \kappa ( s ) \nu ( s ) \ \mathrm d s $$
Note that $ \gamma $ and $ u $ are twice continuously differentiable, which justifies moving the differentiation under the integral. We know that $ \gamma $ is a closed curve; if this is also true for $ u $, the first term in the above expression becomes equal to zero, which implies what you want.
But why did I assume that $ u $ is a closed curve? Because I think that must have been part of the assumptions, but you've missed them or forgotten to mention them in your post. It's not that difficult to construct examples that lack this condition, for which the final claim is false. I suppose you could also find this out thinking about the geometrical intuition behind what the family $ ( \gamma _ t ) _ t $ is: we're trying to smoothly deform the curve $ \gamma $ to another curve; for the resulting curve to be closed, $ u $ must be closed.

Here's a list of what I think is wrong with your post. I suppose these may have misled you somehow, so that you couldn't see how simple the task is.

*

*For the equation $ \tau ' ( s ) = \kappa ( s ) \nu ( s ) $ to hold for the unit tangent vector $ \tau ( s ) $, the (counterclockwise $ \frac \pi 2 $-rotated) unit normal vector $ \nu ( s ) $ and the (signed) curvature $ \kappa ( s ) $, the curve $ \gamma $ must be parametrized with respect to arc length. It's not clear whether you've taken this into account or not. In case you had, you didn't need to write $ \tau ( s ) = \frac { \gamma ' ( s ) } { \| \gamma ' ( s ) \| } $, as it was knows that $ \| \gamma ' ( s ) \| = 1 $, which is not mentioned in your post.

*You've written "$ \psi ( u ) = \frac { \mathrm d } { \mathrm d t } L ( \gamma _ t ) $", suggesting that the expression on the left-hand side of the equation depends not only on $ u $, but also on $ t $, which is not true. In the text you've explained that $ \psi ( u ) $ is the result of differentiating $ L $ with respect to $ t $ at $ t = 0 $. The correct notation would be
$$ \psi ( u ) = \left. \frac { \mathrm d } { \mathrm d t } L ( \gamma _ t ) \right| _ { t = 0 } $$
(in fact, I myself was misled by the notation you had used. I somehow missed the explanation about putting $ t = 0 $, and found out that I have to put $ t = 0 $ through some other route. I originally posted that route in my answer, and after reading the question again, edited my post).

*It seems that you're using the symbol "$ \gamma $" for two different purposes, which is a bad habit, and can make confusions. You've used it to denote a specific curve (which you mention in the beginning of the post, or in definition of the family $ ( \gamma _ t ) _ { t \in \mathbb R } $ of curves), and at the same time as a dummy variable (in the expression "$ L ( \gamma ) = \int _ 0 ^ 1 \| \gamma ' ( s ) \| \ \mathrm d s $", which you intend to later use for a whole class of curves and not just a specific one). Also, because $ u $ is not a fixed curve (for example since it appears as the argument of the functional $ \psi $), a better notation for curves in the family we're interested in would be "$ \gamma ^ u _ t $", so that the dependence on $ u $ becomes explicit. This way, you can also avoid the confusion that in the equation "$ \psi ( u ) = \left. \frac { \mathrm d } { \mathrm d t } L ( \gamma _ t ) \right| _ { t = 0 } $", why doesn't the right-hand side appear to be dependent on $ u $ while the left-hand side appears to be so.

