Why do two definitons of curvature give different answers?

There seems to be only a small mistake in OP's calculation.

We can check the validity of the formula
\begin{align*} \frac{\|T^\prime(t)\|}{\|r^\prime(t)\|}=\frac{\|r^\prime(t)\times r^{\prime\prime}(t)\|}{\|r^\prime(t)\|^3}\tag{1} \end{align*} by taking real-valued functions $x=x(t), y=y(t)$ and setting \begin{align*} r^{\prime}(t)=\langle x(t),y(t)\rangle \end{align*}

We obtain \begin{align*} \|r^\prime(t)\|&=\sqrt{x^2(t)+y^2(t)}\\ T(t)&=\frac{r^\prime(t)}{\|r^\prime(t)\|}\\ &=\left\langle \frac{x(t)}{\sqrt{x^2(t)+y^2(t)}},\frac{y(t)}{\sqrt{x^2(t)+y^2(t)}}\right\rangle\\ T^\prime(t)&=\left\langle\frac{y(t)\left(y(t)x^\prime(t)-x(t)y^\prime(t)\right)}{\left(x^2(t)+y^2(t)\right)^{3/2}},\right.\\ &\left.\qquad\qquad-\frac{x(t)\left(y(t)x^\prime(t)-x(t)y^\prime(t)\right)}{\left(x^2(t)+y^2(t)\right)^{3/2}}\right \rangle\\ &=\frac{y(t)x^\prime(t)-x(t)y^\prime(t)}{\|r^\prime(t)\|^3}\Big\langle y(t),-x(t)\Big\rangle\\ \|T^\prime(t)\|&=\frac{|y(t)x^\prime(t)-x(t)y^\prime(t)|}{\|r^\prime(t)\|^3}\sqrt{x^2(t)+y^2(t)}\\ &=\frac{|y(t)x^\prime(t)-x(t)y^\prime(t)|}{\|r^\prime(t)\|^2}\tag{2}\\ r^\prime(t)\times r^{\prime\prime}(t)&=\langle x(t),y(t),0 \rangle\times\langle x^\prime(t),y^\prime(t),0\rangle\\ &=\langle 0,0,y(t)x^\prime(t)-x(t)y^\prime(t)\rangle\\ \|r^\prime(t)\times r^{\prime\prime}(t)\|&=|y(t)x^\prime(t)-x(t)y^\prime(t)|\tag{3} \end{align*}

From (2) and (3) we obtain \begin{align*} \color{blue}{\frac{\|T^\prime(t)\|}{\|r^\prime(t)\|}}&=\frac{|y(t)x^\prime(t)-x(t)y^\prime(t)|}{\|r^\prime(t)\|^3} \color{blue}{=\frac{\|r^\prime(t)\times r^{\prime\prime}(t)\|}{\|r^\prime(t)\|^3}} \end{align*} and the claim (1) follows.

Now we consider the special case \begin{align*} r^{\prime}(t)&=\langle x(t),y(t)\rangle\\ &=e^t\langle \cos(4t)-4\sin(4t),\sin (4t)+4\cos (4t)\rangle \end{align*}

We calculate \begin{align*} \|r^\prime(t)\|&=\sqrt{x^2(t)+y^2(t)}\\ &=e^t\sqrt{17}\\ r^{\prime\prime}(t)&=e^t\langle -8\sin(4t)-15\cos(4t),8\cos(4t)-15\sin(4t)\rangle\\ \|r^\prime(t)\times r^{\prime\prime}(t)\|&=|y(t)x^\prime(t)-x(t)y^\prime(t)|\\ &=e^{2t} |(\sin(4t)+4\cos(4t))(-8\sin(4t)-15\cos(4t))\\ &\qquad\qquad-(\cos(4t)-4\sin(4t))(8\cos(4t)-15\sin(4t))|\\ &=68e^{2t}\\ \|T^\prime(t)\|&=\frac{68e^{2t}}{17e^{2t}}\\ &=4 \end{align*} We finally obtain \begin{align*} \frac{\|T^\prime(t)\|}{\|r^\prime(t)\|}&\color{blue}{=\frac{4}{e^t\sqrt{17}}}\\ \frac{\|r^\prime(t)\times r^{\prime\prime}(t)\|}{\|r^\prime(t)\|^3}&=\frac{68e^{2t}}{\left(e^t\sqrt{17}\right)^3}\color{blue}{=\frac{4}{e^t\sqrt{17}}} \end{align*}

The problem reduces to a simple miscalculation in end of the second section. In the calculation of the cross product’s norm, I assume there was a calculation error, as it should be $\sqrt{(68e^{2t})^2}$, which becomes $68e^{2t}$. Therefore our second curvature becomes $\frac{4}{\sqrt{17}}e^-t$, which is the same as your curvature by computing it the first way. (In one line in the first part there is a type where you write $e^{-2t}$ however your calculations and your work leading up to it doesn’t use that so I will assume it is a mistake).

In general, both formulas for curvature will hold, and be equal. The only time I can think of would be when it is not a curve embedded in $R^3$ in order to use the cross product, however if you use the formula provided by Sobi you can avoid this problem. The derivation of the cross product definition of curvature begins with the classic definition, so one will never work without the other.