Differentiating composition of functions proof help 
Theorem: Let $X, Y, Z$ be normed spaces and $U\subset X$, $V \subset Y$ open sets. If the function $f:U \to V$ is differentiable in $x \in U$ and function $g: V \to Z$ differentiable in $f(x)\in V$, then the function $g \circ f: U \to Z$ is differentiable in $x \in U,$ and : $$(g \circ f)'(x)=g'(f(x))\circ f'(x)$$

Proof:(I will highlight from which point on is unclear to me..)
-since the functions are differentiable then:
$f(x+h)-f(x)=f'(x)h+R_1(h), \\g(y+h)-g(y)=g'(x)h+R_2(k) $ where $\frac{\|R_1(h)\|}{\|h\|}\to 0 $ when $h\to 0, \frac{\|R_2(k)\|}{\|k\|} \to 0$ when $k \to 0$. Let's give $f(x)$ the notation $y$, when we have :
$$(g \circ f)(x+h)-(g  \circ f )(x)= g(f(x+h))-g(f(x))\\ g(y+f'(x)h+R_1(h))-g(y)=$$

$$g'(y)(f'(x)h)+g'(y)(R_1(h))+R_2(f'(x)h+R_1(h))= \\ (g'(y) \circ f'(x))h+ R(h) \text{ where } R(h)=g'(y)R_1(h)+R_2(f'(x))h + R_1(h).\text { Lets give the vector  } f'(x)h+R_1(h) \text{ the notation } H. \\ \|g'(y)R_1(h)\|\leq \|g'(y)\| \|R_1(h)\|, \\ \|H\| \leq \|f'(x) \|\|h\|+ \| R_1(h)\|, \text{Since }R_1(h)=o(h), R_2(H)=o(H), \text{ from (1),(2),(3) we have:}\\ g'(y)R_1(h)=o(h) \\  H\to 0 \text{ when } h \to 0 \text{ where } \frac{\|H\|}{\|h\|} \text{ is bounded (the professor puts a lot of attention on this part. )}$$

My definition of differentiability:
Let $X$ and $Y$ be normed vector spaces upon the same field $\mathbb R$ or $\mathbb C$ and $U$ an open set in $X$. For a function $f:U \to Y$ it is said to be differentiable in point $x \in U$ if there exists a continuous linear map $A_x:X \to Y$ such that:
$$f(x+h)-f(x)=A_xh+R(h)$$ where $$\lim_{h \to 0}\frac{R(h)}{\|h\|}=0. \text{ or } R(h)=o(h)$$
 A: $
\let\oldcdot\cdot
\renewcommand{\cdot}{\!\oldcdot\!}
\newcommand{\eatLabel}[2]{}
\newenvironment{subequations}{\eatLabel}{}
$
Denoting  $ y:= f(x)$ we have
$$
\begin{subequations}
\label{1}
\begin{align} \label{1:1} \tag{1}
f(x+h) & = y + f'(x) \cdot h+R_1(h)  \\
 g(y+h)& = g(y) +  g'(y) \cdot h+R_2(h)   \label{1:2} \tag{2}
\end{align}
\end{subequations}
$$
Therefore
$$
\begin{aligned}
\big(g \circ f\big)(x+h) -\big(g \circ f\big)(x) 
& = g\big(f(x+h)\big)-g\big(f(x)\big)=\qquad\big\{\text{by }\eqref{1:1}\big\} \\ 
& = g\left(y+f'(x) \cdot h+R_1(h)\right)-g(y)
\end{aligned}
$$
Denoting $H : = f'(x) \cdot h+R_1(h)$, we can write
$$
\begin{aligned}
\big(g \circ f\big)(x+h) -\big(g \circ f\big)(x) 
& = g\left(y+f'(x) \cdot h+R_1(h)\right)-g(y) 
= \\ & = g\left(y+H\right)-g(y) 
\end{aligned}
$$
If we show that $\left \| H\right\| \to 0 $ as soon as $\left\| h\right\| \to 0$, we will be able to  use formula $\eqref{1:2}$.

 Indeed, $\left \| H\right\| =  \left \| f'(x) \cdot h+R_1(h) \right\| $.
 By triangle inequality and Cauchy–Schwarz inequality,
 $\left \| H\right\|\le \left \| f'(x)\right\| \cdot \left \|h\right\|+\left \|R_1(h) \right\| $.
 By your definition of differentiability, $\ f'(y) $ is linear and continuous map, therefore it is bounded: $ \left \| f'(y)\right\|  < \infty$. 
 Since $\left \| R_1(h) \right\|=o(h)$, we get
 $$   \left\| H \right\| = \left\| f'(x) \cdot h+R_1(h)\right\| \le \left\| f'(x) \right\| \cdot \left\| h\right\| +\left\| R_1(h)\right\|  = \left\| f'(x) \right\| \cdot \left\| h\right\| + o(h),\qquad \quad \label{***}\tag{3}$$
so that $\left \| H\right\| \to 0$ as $h\to 0$.

Then, by $\eqref{1:2}$, $g\left(y+H\right)-g(y)  = g'(y)  \cdot H + R_2(H)$, and thus
$$
\big(g \circ f\big)(x+h) -\big(g \circ f\big)(x) = 
g'\big(y\big)  \cdot H + R_2\big(H\big)
$$
Substituting back $H = f'(x) \cdot h+R_1(h) $, we get
$$
\begin{aligned}
\big(g \circ f\big)(x+h) -\big(g \circ f\big)(x) 
& = \dots = g'\big(y\big)  \cdot H + R_2\big(H\big) = 
\\ 
& = g'\big(y\big) \cdot \Big(f'(x) \cdot h+R_1(h)\Big) - 
R_2\Big(f'(x) \cdot h + R_1(h) \Big) 
\\ 
& = g'\big(y\big) \cdot \Big(f'(x) \cdot h\Big) + g'\big(y\big) \cdot \Big( R_1(h)\Big) - R_2\Big(f'(x) \cdot h + R_1(h) \Big) 
\\
& = g'\big(y\big) \cdot f'(x) \cdot h  + g'\big(y\big) \cdot R_1(h) - R_2\Big(f'(x) \cdot h + R_1(h) \Big) 
\\
& = g'\big(y\big)\cdot f'(x)\cdot h+g'\big(y\big)\cdot R_1(h)-R_2\big(H\big) 
\end{aligned}
$$
Replacing $y$ with $f(x)$ and denoting
$R(h):=  g'(y)R_1(h)+R_2(H)=  g'\big(f(x)\big)R_1(h)+R_2(H)$, we get
$$
\boxed{\ 
\big(g\circ f\big)(x+h)-\big(g\circ f\big)(x)=g'\big(\,f(x)\big)\cdot f'(x)\cdot h+R(h) \ \ }
$$

Observe  that $\left\| R(h) \right\| = o(h)$.

 Indeed, since $\left\|R_1(h)\right\|=o(h)$ and $\left\|R_2(h)\right\|=o(h)$, we write
 $$ \begin{aligned} \left\| R(h) \right\| &  =  \left\|g'(y)R_1(h)+R_2(H) \right\| \le \quad   \qquad \  \big\{\text{ by Cauchy–Schwarz and triangle inequalities }\big\} \\ & \le \left\|g'(y)\right\|\left\|R_1(h)\right\|+\left\|R_2(H)\right\|   = \\ & = \left\|g'(y)\right\| o(h) + o(H) =  \quad \quad \qquad   \big\{\text{ by } \eqref{***}\big\}  \\   & = \left\|g'(y)\right\| o(h) + o\big(\!\!\left\| f'(x) \right\| \cdot \left\| h\right\| + o(h)\big) =  \\ & = \Big(\!\!\left\|g'(y)\right\| +\left\|f'(y)\right\|\! \Big) o(h) + o\big( o(h)\big) = o(h), \end{aligned} $$
 and so $\left\|R(h)\right\|=o(h)$.

Then, by definition of differentiability, we write
$$
\begin{cases}
\big( g \circ f \big)'(x) = g'(y)\cdot f'(x)\cdot h+R(h), \\
\left\| R(h) \right\| = o(h), 
\end{cases}
\implies
\bbox[5px, border:3px solid #FF0000]{\  \color{black}{\big( g \circ f \big)'(x) = g'\big(\,f(x)\big)\cdot f'(x) }\ \ }
$$
Q.E.D.
