The question is how define the norm of nth covariant derivative of smooth function f on a manifold M. The manifold is two dimensional so maybe I can do it in the following way: thing about nth covariant derivative as 2^n dimensional vector and take it L^p norm??
If $A$ is any section of a vector bundle $E$ over a smooth manifold $M$ and if $\nabla$ is any covariant derivative on $E$, then $\nabla A$ is a section of $T^*M \otimes E$, and this has a natural (pointwise) inner product $g \otimes h$ given a Riemannian metric $g$ on $M$ and a fibre metric $h$ on $E$, defined by $\langle \alpha \otimes s , \beta \otimes t \rangle_{g \otimes h} = g(\alpha, \beta) h(s, t)$. Now, if you want some kind of global norm, one can take this pointwise norm and do a number of things: if $M$ is compact and oriented (and sometimes even if not), one can take $L^p$ norms by $ \nabla A ^p_{L^P} = \int_M  \nabla A ^p_{g \otimes h} \mathrm{vol_g}$, where $\mathrm{vol_g}$ is the volume form associated to the metric $g$ and chosen orientation. One can also consider $C^k$ norms, Holder norms, and various others. For your specific question, $\nabla^n f$ is a section of $\otimes^n T^* M$, and thus one only needs a metric $g$ on $M$ to define the pointwise norm. 


From the perspective of analysis it is usually more convenient to define your favorite class of function spaces via local coordinates and a partition of unity. More precisely consider your favorite (compact manifold) and choose a covering $U_i$ together with charts $\phi_i:B_1\rightarrow U_i,$ where $B_1$ denotes the unit ball in your model space and such that $\phi$ extends to the closures, and a partition of unity $(\chi_i)$ subordinate to your covering. The you define $\f\_{W^{k,p}(M)}:=\sum_i \\phi_i^\ast (\chi_if)\_{W^{k,p}(B_1)}$ for $f\in C^\infty(M).$ This norm is independent of the choices as long as $M$ is compact and its closure is a Banach space. Vector bundles can be treated similarly. Of course this defines a norm which equivalent to choosing a metric and a connection $\nabla$ and to consider $\f\:=\\nabla^k f\_{L^p}+\f\_{L^p},$ as in Spiro's answer. 

