To prove $H^1(\bar\gamma)\le L$, begin by picking a partition $t_0,\dots, t_m$ such that
$$
\sum\limits_{i=1}^md(\gamma(t_{i-1}),\gamma(t_i)) > L-\epsilon
\tag{1}$$
and $d(\gamma(t_{i-1}),\gamma(t_i))<\epsilon$ for each $i$. Let $A_i = \gamma([t_{i-1},t_i])$.
Suppose $\operatorname{diam} A_i>2\epsilon$ for some $i$. Then there are $t',t''\in (t_{i-1},t_i)$ such that $d(\gamma(t'),\gamma(t''))>2\epsilon$. So, after these numbers are inserted into the partition, the sum of differences $d(\gamma(t_{i-1}),\gamma(t_i)) $ increases by more than $\epsilon$, contradicting $(1)$. Conclusion: $\operatorname{diam} A_i\le 2\epsilon$ for all $i$.
Suppose $\sum_i\operatorname{diam} A_i>L+ \epsilon$. For each $i$ there are $t_i',t_i''\in (t_{i-1},t_i)$ such that $d(\gamma(t'),\gamma(t''))>\operatorname{diam} A_i - \epsilon/m$. So, after all these numbers are inserted into the partition, the sum of differences $d(\gamma(t_{i-1}),\gamma(t_i)) $ will be strictly greater than $L+\epsilon - \epsilon = L$, which is again a contradiction.
Thus, the sets $A_i$ provide a cover such that $\operatorname{diam} A_i\le 2\epsilon$ for all $i$ and $\sum_i\operatorname{diam} A_i\le L+ \epsilon$. Since $\epsilon$ was arbitrarily small, $H^1(\bar\gamma)\le L$.
For completeness: the opposite direction follows from the inequality $$H^1(E)\ge \operatorname{diam} E\tag{2}$$ which holds for any connected set $E$. To prove it, fix a point $a\in E$ and observe that the image of $E$ under the $1$-Lipschitz map $x\mapsto d(x,a)$ is an interval of length close to $\operatorname{diam} E$ provided that $a$ was suitably chosen.
Then apply $(2)$ to each $\gamma([t_{i-1},t_i])$ separately.
It is best to setup the coordinate system in such a manner that the part of curve with length $l_i$ starts from origin and then we have $$l_i=\int_{0}^{h}\sqrt{\{x'(t)\}^2+\{y'(t)\}^2}\,dt$$ and $$d_i=\sqrt{x^2(h)+y^2(h)}$$ and we need to show that $l_i/d_i\to 1$ as $h\to 0$. If we assume that one of the derivatives $x'(0),y'(0)$ is non-zero then we can see via Fundamental Theorem of Calculus that $$\frac{l_i}{h}\to\sqrt{\{x'(0)\}^2+\{y'(0)\}^2}$$ and by definition of derivative $d_i/h$ also tends to the same value and thus our job is done.
The definition of a curve assumes that $x(t), y(t) $ are continuous and further the notion of arc-length requires that these functions be of bounded variation. The corresponding analysis under such general conditions is more difficult.
Based on feedback received in comments let me just mention the development of arc-length and its representation as an integral in brief.
We first begin with the notion of a partition. Let $[a, b] $ be a closed interval. A partition of $[a, b] $ is a finite set $$P=\{t_0,t_1,t_2,\dots,t_n\} $$ such that $$a=t_0<t_1<t_2<\dots<t_n=b$$ Let the set of all possible partitions of $[a, b] $ be denoted by $\mathcal {P} [a, b] $ ie $$\mathcal{P} [a, b] =\{P\mid P\text{ is a partition of }[a, b] \} $$
Let $f:[a, b] \to\mathbb {R} $ be a function. Let $$P=\{t_0,t_1,t_2,\dots,t_n\}$$ be a partition of $[a, b] $ and we form a sum $$V_{f} (P) =\sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|$$ The function $f$ is said to be of bounded variation if the set $$\{V_{f} (P) \mid P\in\mathcal{P} [a, b] \} $$ of sums $V_{f} (P) $ for all partitions $P$ of $[a, b] $ is bounded. And in this case the total variation of $f$ on $[a, b] $ is defined to be the supremum $$V_{f} [a, b] =\sup\, \{V_{f} (P) \mid P\in\mathcal{P} [a, b] \} $$
It is easy to prove that any monotone function is of bounded variation and a bit more difficult to prove that a function is of bounded variation if and only if it can be expressed as a difference of two increasing functions.
Next we come to the topic of interest. Let $f, g$ be two functions from interval $[a, b] $ to $\mathbb {R} $ and let's assume that they are continuous on $[a, b] $. A curve is a set of points $$\mathcal{C} =\{(x, y) \mid x=f(t), y=g(t), t\in[a, b] \} $$ To define the arc-length of this curve $\mathcal{C} $ we start with a partition $$P=\{t_0,t_1,t_2,\dots,t_n\} $$ of $[a, b] $ and form the sum $$L_{\mathcal{C}} (P) =\sum_{i=1}^{n}\sqrt {\{f(t_i)-f(t_{i-1})\}^2+\{g(t_i)-g(t_{i-1})\}^2} $$ In other words corresponding to the partition $P$ of $[a, b] $ we have points $A_0,A_1,A_2\dots,A_n$ on the curve $\mathcal{C} $ with $A_i=(f(t_i), g(t_i)) $ and the above expression for $L_{\mathcal{C}} (P) $ is the sum of lengths of these line segments $A_{i-1}A_{i}$.
If the set $$\{L_{\mathcal{C}} (P) \mid P\in\mathcal{P} [a, b] \} $$ of sums $L_{\mathcal{C}} (P) $ is bounded for all partitions $P$ of $[a, b] $ then we say that the curve $\mathcal{C} $ is rectifiable (ie possesses a well defined arc-length) and its arc-length $L_{\mathcal{C}} $ is defined to be the supremum of all such sums ie $$L_{\mathcal{C}} =\sup\, \{L_{\mathcal{C}} (P) \mid P\in\mathcal{P} [a, b] \} $$ It can be proved with some effort that the curve $\mathcal{C} $ is rectifiable if and only if both $f, g$ are of bounded variation on $[a, b] $.
Let us now assume that the functions $f, g$ used to define the curve $\mathcal{C} $ are differentiable and further $f'(t) \neq 0$ for all $t\in[a, b] $. Then $f$ is one-one (by Rolle's theorem if $f$ takes same value at two points then its derivative vanishes somewhere in between) and for any partition $P=\{t_0,t_1,\dots,t_n\} $ of $[a, b] $ we have $f(t_i) - f(t_{i-1}) \neq 0$. Hence we can write $$L_{\mathcal{C}} (P) =\sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|\sqrt{1+\left(\frac{g(t_i)-g(t_{i-1})}{f(t_i)-f(t_{i-1})}\right)^2}$$ Using Cauchy mean value theorem the ratio inside square roots can be written as $g'(\xi_i) /f'(\xi_i) $ for some $\xi_i\in(t_{i-1}, t_i) $. Also using mean value theorem the expression outside square roots can be written as $|f'(\eta_i)|(t_i-t_{i-1})$ for some $\eta_i\in(t_{i-1},t_i)$ and thus we have $$L_{\mathcal{C}} (P) =\sum_{i=1}^{n}|f'(\eta_i)|\sqrt{1+\left(\frac{g'(\xi_i)}{f'(\xi_i)}\right)^2}(t_i-t_{i-1})$$ and the above looks like a Riemann sum for the integral $$\int_{a} ^{b} |f'(t) |\sqrt{1+\left(\frac{g'(t)}{f'(t)}\right)^2}\,dt=\int_{a}^{b}\sqrt{\{f'(t)\}^2+\{g'(t)\}^2}\,dt$$ (there is a slight technicality involved here due to the different set of points $\eta_i, \xi_i $). As partitions $P$ become finer and finer the expression $L_{\mathcal{C}} (P) $ tends to its supremum $L_{\mathcal{C}} $ and the Riemann sums above tend to the integral above and thus we get the arc-length formula $$L_{\mathcal{C}} =\int_{a}^{b}\sqrt{\{f'(t)\}^2+\{g'(t)\}^2}\,dt$$ This assumes that the integral on right exists.
In the above development we have nowhere use the fact $s_i/d_i\to 1$. The integral formula for arc-length is a consequence of definition of arc-length and mean value theorems. And the result in your question is a consequence of this integral formula.
The restriction $f'(t) \neq 0$ can be removed by using Langrange Mean Value Theorem on $f, g$ in the sum $L_{\mathcal {C}} (P) $ and Duhamel Principle for integrals.
Best Answer
I am not sure this is what the author intended, but the claim that holds "by definition" can be shown as follows:
Take any $t_1<t_0$. Then $L(C|[t_1,t_0])\ge \epsilon_0$, so there exists a partition $t_1=s_0 < \cdots < s_n=t_0$ such that $$\sum_{j=1}^n |\phi(s_j)-\phi(s_{j-1})|> 3 \epsilon_0/4.$$ We may assume $|\phi(s_n)-\phi(s_{n-1})|< \epsilon_0/4$, since $\phi$ is continuous.
(Otherwise there exists $\hat s_n \in (s_{n-1},t_0)$ with $|t_0-\hat s_n|<1/4 \epsilon_0$. Take $\hat s_j = s_j$ for $j=1,\ldots n-1$, and $\hat s_{n+1}=t_0$. Then $$\sum_{j=1}^{n+1} |\phi(\hat s_j)-\phi(\hat s_{j-1})| \ge \sum_{j=1}^n |\phi(s_j)-\phi(s_{j-1})|> 3 \epsilon_0/4$$ by triangle inequality.)
Hence $L(C|[t_1,s_{n-1}])>\epsilon_0/2$. Set $t_2=s_{n-1}$.
We still have $L(C|[t_2,t_0])\ge\epsilon_0$, so we can start over to get $t_3, t_4,\ldots$ with $L(C|[t_j,t_{j+1}]) > \epsilon_0/2$ by induction.
(Not sure why we need $\tilde t_j$, but if we want to, we can take any $\tilde t_j \in (t_j,t_0)$ and then proceed in step 2 with $\tilde t_j$ instead of $t_j$.)