Explaination of the Frechet Derivative 
Let $E$ and $F$ be Banach spaces and let $U$ be an open subset of $E$. Suppose $g:U \to F $. $g$ is continuous at $x_0$ if there exists a linear transformation, $T_{x_0}$, such that $$ \lim \limits_{x_0 \to 0}\dfrac{||g(x_0+h)-g(x_0)-T_{x_0}(h)||}{||h||}. $$ Then $Dg_{x_0}=T_{x_0}$ is then the Frechet derivative of $g$ at $x_0$.

I have a question about this with two examples from this video (https://www.youtube.com/watch?v=RKHx1vQdZko).
For the first example we compute the Frechet derivative of $f(x)=x^2$ and we get that $T_x(h) = h(2x)$. For the second example we find the Frechet derivative of $f(x,y)=(xy , x+y)$ and here we get that $T_{(x,y)}(h,v)=\begin{bmatrix}
    y       & x \\
    1    &   1  \\  
\end{bmatrix}\begin{bmatrix}
    h       \\
    v    \\  
\end{bmatrix}$ .
But the definition says that $Df_{x_0}=T_{x_0}$ so does that mean that $Df_{x_0}=T_{x_0}(1)$? It seems to be the case for both of these examples. I believe this is always but am unsure. The reason why I think this is that the transformation is linear so $T_{x_0}(h)=h(Df_{x_0})$ since $T$ linearly maps $h$ so if we let $h=1$ then we are just left with $Df_{x_0}$. Does my reasoning make sense? 
 A: Yes, it makes sense in case $E = \mathbb{K}$ where $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$, i.e. what you said makes sense for your first example. The comment of @WillM. is key here, as $h=1$ in Banach spaces only means something if the Banach space is a field too. Note that your equation $T_{x_0}(h) = h(Df_{x_0})$ doesn't make sense in any other case.
More precisely: $\mathcal{L}(\mathbb{K},F)$ is isometrically isomorphic to $F$ via $\varphi\colon\mathcal{L}(\mathbb{K}, F)\to F,\, A\mapsto A1$.
To see this, note that $\varphi$ obviously is linear and injective. Next, consider, given $v\in F$, the definition $(A_vx:=x\cdot v,\; x\in\mathbb{K})$. That way we get $A_v\in\mathcal{L}(\mathbb{K}, F)$ and $A_v1 = v$, i.e. $\varphi$ is surjective. Finally, 
$$ \Vert Ax\Vert_F = \vert x\vert \Vert A1\Vert_F\leq \Vert A1\Vert_f,\quad \vert x\vert \leq 1,\; A\in\mathcal{L}(\mathbb{K}, F),$$
which easily shows that $\Vert A\Vert_{\mathcal{L}(\mathbb{K},F)} = \Vert A1\Vert_F$. 
In conclusion, you may identify $\mathcal{L}(\mathbb{K}, F) = F$.
