Am I forgetting a $2$ in calculating the area of $S=\{(x,y,z)\in\mathbb{R}^3 | x^2+y^2+z^2=25, 2x^2+y^2\leq 25\}$?
Here is my solution: $$z^2+y^2+z^2 = 25 \Rightarrow z = \pm\sqrt{25-x^2-y^2}$$ We can now consider only the top-half of the sphere and double the area at the end. Now I need to calculate $$||T_x\times T_y||=||\begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \newline 1 & 0 & \frac{-x}{\sqrt{25-x^2-y^2}} \newline 0 & 1 & \frac{-y}{\sqrt{25-x^2-y^2}}\end{vmatrix}|| = \sqrt{\frac{25}{25-x^2-y^2}}$$
Finally, I need to calculate $$\int\int_D\sqrt{\frac{25}{25-x^2-y^2}}\text{d}x\text{d}y$$ To do this, I transform into polar coordinates: $$\begin{cases}x=5r\cos(t) \\ y=5r\sin(t)\end{cases}, r\in[0,1], t\in[0,2\pi]$$ From $2x^2+y^2\leq 25 \Rightarrow 25r^2\cos^2(t)+25r^2\leq 25 \Rightarrow r \leq\sqrt{\frac{1}{\cos^2(t)+1}}$.
Thus, the final integral is $$\int_{0}^{2\pi}\int_{0}^{\sqrt{\frac{1}{\cos^2(t)+1}}}\frac{1}{\sqrt{1-r^2}}\cdot 25r\ \text{d}r\text{d}t$$
However, even after doubling the result of this, my result ends up being half of that given in my book (which is $100\pi$), and I am unable to see where I am missing a $2$ along the way. Any help here is much appreciated!
Your answer $(50\pi)$ is correct, and you don't especially need calculus to see why. First note that $100\pi$ is the area of the sphere $x^2+y^2+z^2=25$, too much to be the area you want, which is clearly a subset of the sphere.
The intersection of the sphere and ellipse is a pair of perpendicular planes,
$$\begin{cases}x^2+y^2+z^2=25\\2x^2+y^2=25\end{cases} \implies z^2=x^2$$
and these planes slice the sphere into $4$ uniform pieces of equal area, and $2$ of these would make up $50\pi$.
Let's confirm with another integral. In spherical coordinates $(x,y,z)=(\rho\cos\theta\sin\varphi,\rho\sin\theta\sin\varphi,\rho\cos\varphi)$, restricting $z\ge|x| \iff |\varphi|\le\dfrac\pi4$, the intersection above gives the equation
$$\rho^2 \cos^2\varphi = \rho^2 \cos^2\theta \sin^2\varphi \implies \tan \varphi = \pm \sec \theta$$
so that by symmetry, the area is given by
$$A = 2 \int_0^{2\pi} \int_0^{\arctan(\sec\theta)} 25\sin\varphi \, d\varphi \, d\theta$$
Now recall that $\cos\left(\arctan x\right)=\dfrac1{\sqrt{1+x^2}}$, and take advantage of symmetry to reduce the integral to
$$\begin{align*} A &= 50 \int_0^{2\pi} \left(1 - \frac{1}{\sqrt{1+\sec^2\theta}}\right) \, d\theta \\ &= 100\pi - 50 \int_0^{2\pi} \frac{d\theta}{\sqrt{1+\sec^2\theta}} \\ &= 100\pi - 50\times4 \int_0^\tfrac\pi2 \frac{d\theta}{\sqrt{1+\sec^2\theta}} \\ &= 100\pi - 200 \underbrace{\int_1^\infty \frac{du}{u\sqrt{u^4-1}}}_{=\tfrac\pi4} & u=\sec t \\ &= \boxed{50\pi} \end{align*}$$
The indicated integral is elementary:
$$\int \frac{du}{u \sqrt{u^4-1}} \stackrel{v^2=u^4-1}= \frac12 \int \frac{v}{\lvert v\rvert \left(v^2+1\right)} \, dv$$