Take $s \ge 1$.

Let $C=2^{1+s/2}(1+c)$. I claim: $\mathcal H^s(G) \le C$.

Let $N \in \mathbb N$. Let $\eta > 0$ be so small that $N2^s\eta^s \le c$.

For $j=1,2,\dots$ let $$ M_j = \sup\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\},\qquad m_j = \inf\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\} . $$

Now construct a partition $\mathcal P$ of $[0,1]$. First put $0,1 \in \cal P$. Then, for each $j$ from $1$ to $N$, put $x_j, y_j \in \cal P$, chosen so that $|f(x_j) - M_j| < \eta$ and $|f(y_j) - m_j| < \eta$. Then $|M_j - m_j| \le |f(x_j)-f(y_j)| + 2 \eta$. Estimate \begin{align} \sum_{j=1}^N |M_j-m_j|^s &\le \sum_{j=1}^N\big(|f(x_j)-f(y_j)|+2 \eta\big)^s \\ &\le N2^s\eta^s+\sum_{j=1}^N |f(x_j))-f(y_j)|^s \le N2^s\eta^s+c \le 2c . \end{align}

Now construct a cover $\mathcal Q$ of $G$. Subdivide the plane into squares of side $1/N$ using vertical lines $x=k/N, k \in \mathbb Z$ and horizontal lines $y=k/N, k \in \mathbb Z$. Let $\mathcal Q$ include those squares that meet the graph $G$. Let us say an index $j$ is normal if $M_j - m_j < 1/N$, and abnormal if $M_j - m_m \ge 1/N$

For a given $j$, how many squares in $\mathcal Q$ are there with base $[(j-1)/N,j/N]$? If $j$ is normal, then there are at most $2$. If $j$ is abnormal then there are at most $$ \lceil NM_j\rceil - \lfloor Nm_j\rfloor \le N(M_j-m_j) +2 . $$ Now compute $$ \sum_{U \in \mathcal Q} \text{diam}(U)^s \le \left(\frac{\sqrt2}{N}\right)^s \left[\sum_{j\text{ normal}} 2+ \sum_{j\text{ abnormal}}\big( N(M_j-m_j) +2\big)\right] $$ But $\sum_{j=1}^N 2 = 2N$ and $$ \sum_{j\text{ abnormal}} N(M_j-m_j) \le N \sum_{j\text{ abnormal}} (M_j-m_j)^s N^{s-1} \le N^s \sum_{j=1}^N (M_j - m_j)^s \le 2cN^s . $$ Therefore $$ \sum_{U \in \mathcal Q} \text{diam}(U)^s \le \frac{2^{s/2}}{N^s}\big[2N + 2cN^s\big] =\frac{2^{1+s/2}}{N^{s-1}} + 2^{1+s/2} c \le 2^{1+s/2} (1+c) = C $$

Therefore $\mathcal H^s_{\sqrt{2}/N} \le C$.

This is true for all (large enough) $N$, so $\mathcal H^s(G) \le C$.

Take $s \ge 1$. We will use $(x+y)^s \le x^s+y^s$ for positive $x,y$.

Let $C=2^{1+s/2}(1+c)$. I claim: $\mathcal H^s(G) \le C$.

Let $N \in \mathbb N$. Let $\eta > 0$ be so small that $N2^s\eta^s \le c$.

For $j=1,2,\dots$ let $$ M_j = \sup\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\},\qquad m_j = \inf\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\} . $$

Now construct a partition $\mathcal P$ of $[0,1]$. First put $0,1 \in \cal P$. Then, for each $j$ from $1$ to $N$, put $x_j, y_j \in \cal P$, chosen so that $|f(x_j) - M_j| < \eta$ and $|f(y_j) - m_j| < \eta$. Then $|M_j - m_j| \le |f(x_j)-f(y_j)| + 2 \eta$. This partition has $2N+2$ dividing points in it. Estimate \begin{align} \sum_{j=1}^N |M_j-m_j|^s &\le \sum_{j=1}^N\big(|f(x_j)-f(y_j)|+2 \eta\big)^s \\ &\le N2^s\eta^s+\sum_{j=1}^N |f(x_j))-f(y_j)|^s \le N2^s\eta^s+c \le 2c . \end{align}

Now construct a cover $\mathcal Q$ of $G$. Subdivide the plane into squares of side $1/N$ using vertical lines $x=k/N, k \in \mathbb Z$ and horizontal lines $y=k/N, k \in \mathbb Z$. Let $\mathcal Q$ include those squares that meet the graph $G$. Let us say an index $j$ is normal if $M_j - m_j < 1/N$, and abnormal if $M_j - m_m \ge 1/N$

For a given $j$, how many squares in $\mathcal Q$ are there with base $[(j-1)/N,j/N]$? If $j$ is normal, then there are at most $2$. If $j$ is abnormal then there are at most $$ \lceil NM_j\rceil - \lfloor Nm_j\rfloor \le N(M_j-m_j) +2 . $$ Now compute $$ \sum_{U \in \mathcal Q} \text{diam}(U)^s \le \left(\frac{\sqrt2}{N}\right)^s \left[\sum_{j\text{ normal}} 2+ \sum_{j\text{ abnormal}}\big( N(M_j-m_j) +2\big)\right] $$ But $\sum_{j=1}^N 2 = 2N$ and $$ \sum_{j\text{ abnormal}} N(M_j-m_j) \le N \sum_{j\text{ abnormal}} (M_j-m_j)^s N^{s-1} \le N^s \sum_{j=1}^N (M_j - m_j)^s \le 2cN^s . $$ Therefore $$ \sum_{U \in \mathcal Q} \text{diam}(U)^s \le \frac{2^{s/2}}{N^s}\big[2N + 2cN^s\big] =\frac{2^{1+s/2}}{N^{s-1}} + 2^{1+s/2} c \le 2^{1+s/2} (1+c) = C $$

Hence $\mathcal H^s_{\sqrt{2}/N} \le C$.

This is true for all (large enough) $N$, so $\mathcal H^s(G) \le C$.

Take $s \ge 1$, since only constant functions satisfy our condition if $s<1$. We will use the inequality $(x+y)^s \le x^s+y^s$ for positive $x,y$. Also take $s \le 2$ since if $s>2$ then $H^s(G) = 0$ for all subsets $G$ of the plane.

Let $C=2^{1+s/2}(1+c)$ I claim: $H^s(G) \le C$.
Fix $\epsilon > 0$. It suffices to find a oountable cover $\{U_j\}_{j \in \mathbb N}$ of $G$ so that $\sum_j \text{diam}(U_j)^s \le C$.

Let $N \in \mathbb N$ be so large that $\sqrt{2}/N < \epsilon$. Let $\eta > 0$ be so small that $N2^s\eta^s \le c$.

For $j=1,2,\dots$ let $$ M_j = \sup\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\},\qquad m_j = \inf\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\} . $$

Now construct a partition $\mathcal P$ of $[0,1]$. First put $0,1 \in \cal P$. Then, for each $j$ from $1$ to $N$, put $x_j, y_j \in \cal P$, where $|f(x_j) - M_j| < \eta$ and $|f(y_j) - m_j| < \eta$. Then $|M_j - m_j| \le |f(x_j)-f(y_j)| + 2 \eta$. \begin{align} \sum_{j=1}^N |M_j-m_j|^s &\le \sum_{j=1}^N\big(|f(x_j)-f(y)j)|+2 \eta\big)^s \\ &\le N2^s\eta^s+\sum_{j=1}^N |f(x_j))-f(y_j)|^s \le N2^s\eta^s+c \le 2c \tag{1}\end{align}

Now construct a cover $\mathcal Q$ of $G$. Subdivide the plane into squares of side $1/N$ using vertical lines $x=k/N, k \in \mathbb Z$ and horizontal lines $y=k/N, k \in \mathbb Z$. Let $\mathcal Q$ include those squares that meet the graph $G$. Let us say an index $j$ is normal if $M_j - m_j < 1/N$, and abnormal if $M_j - m_m \ge 1/N$

For a given $j$, how many squares in $\mathcal Q$ are there with base $[(j-1)/N,j/N]$? If $j$ is normal, then there are at most $2$. If $j$ is abnormal then there are at most $$ \lceil NM_j\rceil - \lfloor Nm_j\rfloor \le N(M_j+m_j) +2 . $$ Now compute $$ \sum_{U \in \mathcal Q} \text{diam}(U)^s \le \left(\frac{\sqrt2}{N}\right)^s \left[\sum_{j\text{ normal}} 2+ \sum_{j\text{ abnormal}}\big( N(M_j+m_j) +2\big)\right] $$ Now $\sum_{j=1}^N 2 = 2N$ and $$ \sum_{j\text{ abnormal}} \big( N(M_j+m_j)\big) \le N \sum_{j\text{ abnormal}} (M_j+m_j)^s N^{s-1} \le N^s \sum_{j=1}^N (M_j + m_j)^s \le 2cN^s . $$ Therefore $$ \sum_{U \in \mathcal Q} \text{diam}(U)^s \le \frac{2^{s/2}}{N^s}\big[2N + 2cN^s\big] =\frac{2^{1+s/2}}{N^{s-1}} + 2^{1+s/2} c \le 2^{1+s/2} (1+c) = C $$

Therefore $H^s_{\sqrt{2}/N} \le C$.

This is true for all (large enough) $N$, so $H^s(G) \le C$.

Take $s \ge 1$.

Let $C=2^{1+s/2}(1+c)$. I claim: $\mathcal H^s(G) \le C$.

Let $N \in \mathbb N$. Let $\eta > 0$ be so small that $N2^s\eta^s \le c$.

For $j=1,2,\dots$ let $$ M_j = \sup\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\},\qquad m_j = \inf\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\} . $$

Now construct a partition $\mathcal P$ of $[0,1]$. First put $0,1 \in \cal P$. Then, for each $j$ from $1$ to $N$, put $x_j, y_j \in \cal P$, chosen so that $|f(x_j) - M_j| < \eta$ and $|f(y_j) - m_j| < \eta$. Then $|M_j - m_j| \le |f(x_j)-f(y_j)| + 2 \eta$. Estimate \begin{align} \sum_{j=1}^N |M_j-m_j|^s &\le \sum_{j=1}^N\big(|f(x_j)-f(y_j)|+2 \eta\big)^s \\ &\le N2^s\eta^s+\sum_{j=1}^N |f(x_j))-f(y_j)|^s \le N2^s\eta^s+c \le 2c . \end{align}

Now construct a cover $\mathcal Q$ of $G$. Subdivide the plane into squares of side $1/N$ using vertical lines $x=k/N, k \in \mathbb Z$ and horizontal lines $y=k/N, k \in \mathbb Z$. Let $\mathcal Q$ include those squares that meet the graph $G$. Let us say an index $j$ is normal if $M_j - m_j < 1/N$, and abnormal if $M_j - m_m \ge 1/N$

For a given $j$, how many squares in $\mathcal Q$ are there with base $[(j-1)/N,j/N]$? If $j$ is normal, then there are at most $2$. If $j$ is abnormal then there are at most $$ \lceil NM_j\rceil - \lfloor Nm_j\rfloor \le N(M_j-m_j) +2 . $$ Now compute $$ \sum_{U \in \mathcal Q} \text{diam}(U)^s \le \left(\frac{\sqrt2}{N}\right)^s \left[\sum_{j\text{ normal}} 2+ \sum_{j\text{ abnormal}}\big( N(M_j-m_j) +2\big)\right] $$ But $\sum_{j=1}^N 2 = 2N$ and $$ \sum_{j\text{ abnormal}} N(M_j-m_j) \le N \sum_{j\text{ abnormal}} (M_j-m_j)^s N^{s-1} \le N^s \sum_{j=1}^N (M_j - m_j)^s \le 2cN^s . $$ Therefore $$ \sum_{U \in \mathcal Q} \text{diam}(U)^s \le \frac{2^{s/2}}{N^s}\big[2N + 2cN^s\big] =\frac{2^{1+s/2}}{N^{s-1}} + 2^{1+s/2} c \le 2^{1+s/2} (1+c) = C $$

Therefore $\mathcal H^s_{\sqrt{2}/N} \le C$.

This is true for all (large enough) $N$, so $\mathcal H^s(G) \le C$.