Addressing the first part of the question requires application of the [change of variables theorem](https://mathworld.wolfram.com/ChangeofVariablesTheorem.html). We denote the twice differentiable $C^2(\mathbb{R})$ (for convenience) functions $x: t \mapsto x(t), v: t \mapsto \frac{dx}{dt}(t)$, $a: t \mapsto \frac{d^2x}{dt^2}(t)$ and $F_c + F_{nc}: x \mapsto F_c(x) + F_{nc}(x)$, where Newton's second law of motion implies that $a(x) = \frac{1}{m}(F_c + F_{nc})(x)$. The definite integral manipulation which uses the change of variable $x = x(t)$, given as 
$$\frac{1}{m} \int_{x_1}^{x_2} (F_c + F_{nc})(x) \; dx = \int_{x_1}^{x_2} a(x) \; dx = \int_{t_1}^{t_2} a(x(t)) \frac{dx}{dt} \; dt,$$ where $x(t_1) = x_1$ and $x(t_2) = x_2$ implies that 
$$\int_{x_1}^{x_2} a(x) \; dx = \int_{t_1}^{t_2} \frac{dv}{dt}(t) v(t) \; dt,$$ while the change of variable $v = v(t)$ (applied *in reverse*, to state the manipulation in words) implies that 
$$\int_{x_1}^{x_2} a(x) \; dx = \int_{t_1}^{t_2} \frac{dv}{dt}(t(v)) v(t(v))  \; dv = \int_{v_1}^{v_2} v(v) \; dv = {\Huge[} \frac{v^2}{2}{\Huge]}{\Huge|}_{v_1}^{v_2},$$
where $v(t_1) = v_1$ and $v(t_2) = v_2$. This completes the first part of the answer.

The vector dot product $(a_x \hat{x} + a_y \hat{y} + a_z \hat{z}) \cdot (d\vec{r}) = a_x d x + a_y dy + a_z dz$ so that, using the previous result, $$\int_{C} (a_x \hat{x} + a_y \hat{y} + a_z \hat{z}) \cdot (d\vec{r}) = \int_{(x_1,y_1,z_1)}^{(x_2,y_2,z_2)} (a_x \hat{x} + a_y \hat{y} + a_z \hat{z}) \cdot (d\vec{r}) \\ = \int_{x_1}^{x_2} a_x dx + \int_{y_1}^{y_2} a_y dy + \int_{z_1}^{z_2} a_z dz = \frac{1}{2}({\large[}v_x^2{\large]}{\Huge|}_{v_{x1}}^{v_{x2}} + {\large[}v_y^2{\large]}{\Huge|}_{v_{y1}}^{v_{y2}} + {\large[}v_z^2{\large]}){\Huge|}_{v_{z1}}^{v_{z2}} = \frac{1}{2} [\|\vec{v}\|^2]{\Huge|}_{\vec{v}_1}^{\vec{v}_2},$$ which completes the second part of the answer.