why is $W^{k,p}(\Sigma, u^*TM)$ the tangent space of $W^{k,p}(\Sigma,M)$ at $u$ Let $\Sigma$ and $M$ be two smooth manifolds with $\Sigma$ compact.
It is a well-known fact that $W^{k,p}(\Sigma, u^*TM)$ is the tangent space of $W^{k,p}(\Sigma,M)$ at $u$. (You can replace these spaces by $C^\infty(\Sigma, u^*TM)$ and $C^\infty(\Sigma, M)$)
But when someone asked me to make a clear and detailed explanation, I was not able to do so. So, does anybody have a concise but rigorous interpretation of this fact? Thanks.
 A: This follows from the construction of the charts.
The case $C^k(M,N)$: For $M$ a compact manifold, $N$ a Riemannian manifold and $k\in \mathbb{N}$ one can construct charts for the smooth Banach manifold $C^k(M,N)$ as follows: given $f\in C^k(M,N)$ there exists a neighborhood $U_f\subset C^k(M,N)$ of $f$ (in the $C^k$-compact open topology) s.t. the maps
$\varphi\colon U_f\to \Gamma_{C^k}(f^*TN),\hspace{3em} g\mapsto (x\mapsto {exp}_{f(x)}^{-1}g(x))$
(where exp denotes the Riemannian exponential map of $N$) are homeomorphisms onto their images. One can show that the chart changes are smooth. From this it follows that the model of the tangent space of $C^k(M,N)$ at $f$ is $\Gamma_{C^k}(f^*TN)$.
The case $k=\infty$ can be found in The Convenient Setting of Global Analysis (Peter Michor & Andreas Kriegel) Chapter IX, Manifolds of Mappings.
For the case of Sobolev spaces you can take a look at Calculus of Variations and Harmonic Maps (Hajime Urakawa), Theorem 4.33 on page 73.
A: Start with the simplest case $M=\mathbb R^n$. Then $W^{k,p}(\Sigma, \mathbb R^n)$ is vector space and the tangent space is trivially
$$ T_u W^{k,p}(\Sigma, \mathbb R^n) = W^{k,p}(\Sigma, \mathbb R^n)\,.$$
You can identify this with the sections of the pullback bundle, $\Gamma_{W^{k,p}}(\Sigma \leftarrow u^\ast T \mathbb R^n)$. Intuitively this means that an infinitesimal deformation of a map is a vector field along this map with values in the ambient tangent space.
For $M$ a manifold, one can embed $M$ into some Euclidean space using Whitney's embedding theorem. Then, intuitively an infinitesimal deformation has to be tangent to the manifold $M$. An alternative is to follow the path shown by user97669: choose a Riemannian metric on $M$ and linearize the manifold around $u(\Sigma)$ using the exponential map.
Note that in the general case the tangent space are not all maps
$ W^{k,p}(\Sigma, u^\ast TM) $, but only sections (of the corresponding regulartiy class)
$$ T_u W^{k,p}(\Sigma, M) = \Gamma_{W^{k,p}}(\Sigma \leftarrow u^\ast TM )\,. $$
The whole tangent bundle can be identified with the space
$$ T W^{k,p}(\Sigma, M) = W^{k,p}(\Sigma, TM)\,. $$
A different (but equivalent) differential structure on $W^{k,p}(\Sigma, M)$ when $p=2$ was constructed in On the Regularity of the Composition of Diffeomorphisms, by Inci, Kappeler and Topalov, 2013.
A: Heuristic:
The tangent space at a point is equal to the space of "velocity vectors" of curves. In particular, given any $x \in N$ and $v \in T_xN$, there exists a curve $\gamma:(-1,1)\rightarrow N$, such that $\gamma(0) = x$ and $\gamma'(0) = v$.
So given any $u \in C^\infty(M,N)$ and tangent vector $v \in T_uC^\infty(M,N)$, there is a curve $\gamma: (-1,1)\rightarrow C^\infty(M,N)$ such that $\gamma(0) = u$ and $\gamma'(0) = v$. On the other hand, $\gamma$ is equivalent to the map $\Gamma: (-1,1)\times M \rightarrow N$, where $\Gamma(t,x) = \gamma(t)(x)$. So
$$v(x) = \gamma'(0)(x) = \partial_t\Gamma(0,x) \in T_{\Gamma(0,x)}N = T_{\gamma(0)(x)}N = T_{u(x)}N = (u^*T_*N)_x.$$ Therefore, $v$ is a section of $u^*T_*N$.
