In standard multivariable calculus we consider functions $g : U \to V$ between open subsets $U \subset \mathbb R^m$ and $V \subset \mathbb R^n$. In fact, one can easily see that differentiability in $x \in U$ implies continuity in $x$.
do Carmo defines the concept of differentiability of a function $\varphi : V \to S_2$ by reducing it to the differentiability of the functions $\mathbf{x}_2^{-1}\circ \varphi \circ \mathbf{x}_1:U_1 \to U_2$ in the multivariable sense. In my opinion is not absolutely clear what he means by
given parametrizations
$$\mathbf{x}_1: U_1 \subset \mathbb{R}^2 \to S_1, \quad \mathbf{x}_2: U_2\subset \mathbb{R}^2 \to S_2$$
with $p \in \mathbf{x}_1(U_1)$ and $\varphi(\mathbf{x}_1(U_1))\subset \mathbf{x}_2(U_2)$.
Does he mean that we have to consider some such pair of parametrizations of all such pairs? Anyway, one can show that this does not really matter. In fact, do Carmo writes
The proof that this definiton does not depend on the choice of parametrizations is left as an exercise.
Whatever the "correct" interpretation of the definition may be, the continuity of $\varphi$ at $p$ is needed to assure the existence of a pair of parametrizations such that $\varphi(\mathbf{x}_1(U_1))\subset \mathbf{x}_2(U_2)$. Note that $p \in \mathbf{x}_1(U_1)$ and $\varphi(\mathbf{x}_1(U_1))\subset \mathbf{x}_2(U_2)$ is only possible if $\varphi(p) \in \mathbf{x}_2(U_2)$. That is, the minimal requirement is that we take parameterizations $\mathbf{x}_1$ around $p \in S_1$ and $\mathbf{x}_2$ around $\varphi(p) \in S_2$.
We can of course always consider the subset $\varphi^{-1}(\mathbf{x}_2(U_2)) \subset S_1$, but if $\varphi$ is not continuous at $p$, then we cannot be sure that $\varphi^{-1}(\mathbf{x}_2(U_2))$ is an open neigborhood of $p$ in $S_1$. It is even possible that $\varphi^{-1}(\mathbf{x}_2(U_2)) = \{p\}$. Thus there may not exist a parameterization $\mathbf{x}_1$ around $p \in S_1$ such that $\mathbf{x}_1(U_1) \subset \varphi^{-1}(\mathbf{x}_2(U_2))$ which is the same as $\varphi(\mathbf{x}_1(U_1))\subset \mathbf{x}_2(U_2)$.
You should compare this with Definition 1 on p. 72. Here do Carmo considers functions (no continuity assumption!) $f: V \to \mathbb R$ and defines differentiability via a parameterization around $p$. Continuity is not required here because $f$ maps each open neigborhood of $p$ in $S$ to $\mathbb R$. It is easy to see that differentiability implies continuity in this case.
Update:
Here is an example of "poor behavior" of non-continuous maps.
Take $S_1 = U \times \{0\} \subset \mathbb R^3$ with $U = \{(x,y) \in \mathbb R^2 \mid x^2 + y^2 < 1\}$ and $S_2 = S^2$ = unit sphere in $\mathbb R^3$. Define
$$\varphi: S_1 \to S_2, \varphi(x,y, 0) = \begin{cases} (x,y, \sqrt{1 - x^2-y^2}) & (x,y) \ne (0,0) \\ (0,0,-1) & (x,y) = (0,0) \end{cases}$$
Consider the point $p = (0,0,0) \in S_1$. Around $\varphi(p)$ we take do Carmo's parameterization $\mathbf{x_2} : U \to \mathbb R^3$ defined on p.56. Then $\varphi^{-1}(\mathbf{x}_2(U)) = \{p\}$ and therefore no parameterization $\mathbf{x_1} : U_1 \to \mathbb R^3$ of $S_1$ around $p$ has the property $\varphi(\mathbf{x}_1(U_1))\subset \mathbf{x}_2(U_2)$. You can nevertheless consider the function $\mathbf{x}_2^{-1}\circ \varphi \circ \mathbf{x}_1$, but its domain is only $\{(0,0)\}$. It does not make sense to call such a function differentiable at $q = (0,0)$.
Best Answer
Essentially it is because of the chain rule.
For functions $F \colon M \to N$ and $G \colon N \to K$, the differential of the composite map $G \circ F \colon M \to K$ is $$ d(G \circ F)_m = dG_{F(m)} \circ dF_m $$
In your case, you have $$ (f \circ \gamma)'(0) = df_{\gamma(0)} \circ \gamma'(0) $$
So if $\gamma'(0) = \alpha'(0)$, then you get $(f \circ \alpha)'(0) = (f \circ \gamma)'(0)$.