Does decomposition of motion rely on Pythagorean theorem? As an example, when analyzing a simple projectile motion with initial horizontal velocity in Newtonian mechanics, I'm enabled to decompose the projectile motion into the vertical and horizontal components of velocity. But if I'm only interested in the speed at some time, I can also use work–energy principle to avoid the difficult elaborate calculation.
Note that the kinetic energy is in term of $v^2$ rather than $v$, and the formula for kinetic energy is derived without Pythagorean theorem (which implies the two needn't to be consistent). So if Pythagorean theorem is not in its now form, does kinetic energy no more follows an algebraic summation, and we can no more decompose a motion?
 A: To answer the question in your title, resolution of vectors into unique components depends only (1) on the fact of the vector space over the underlying field; as long as one has a basis (any $N$ linearly independent vectors where $N$ is the space's dimension), unique resolution of a vector into components with respect to this basis is defined.
However, notions of work require the vector space to be an inner product space where the inner product is a bilinear form (a function $\langle\cdot,\,\cdot\rangle:V\times V\to\mathbb{R}$ linear in both its arguments) where $\langle X,\,X\rangle>0$ when $X\neq0$ and $\langle X,\,X\rangle=0\Leftrightarrow X=0$ and, for any such bilinear form, we can always choose a basis where a vector's length is given by the Pythagorean formula. So, in effect, the Pythagorean formula needfully accompanies the structure needed to define work meaningfully. In curved space - a Riemannian manifold - the tangent spaces are all inner product spaces (this is the definition of a Riemannian manifold) and so even there the Pythagorean formula holds locally: we can always diagonalize a nonsignatured metric at any particular point to the Pythagorean form.
A: I have a hard time understanding your question, but energy and power are independent of any transformations performed on the coordinate frame that you use.
Consider a particle with mass $m$ and velocity $\dot{\bar{x}}_1=[\dot{x}_1\ \dot{y}_1]^\top$ in coordinate frame $X_1$. 
Consider a secondary frame $X_2$, aligned with the absolute velocity of the same particle. The velocity of the particle in this frame is $\dot{\bar{x}}_2=[\dot{x}_2\ \dot{y}_2]^\top$ with  $\dot{x}_2 = \sqrt{\dot{x}_1^2 + \dot{y}_1^2}$ and $\dot{y}_2 = 0$.
The kinetic energy is
$$E = \frac{1}{2}m\left(\dot{x}_1^2+ \dot{y}_1^2\right)=\frac{1}{2}m\left(\dot{x}_2^2 + \dot{y}_2^2\right)$$
Fill in the equations and you'll see that they are the same.
