They're both ANOVA; they're both about partitioning the total corrected sum of squares into an explained part and an unexplained part.
Suppose
\begin{align}
U_{ij} = {} & (\text{average for group }i) + (\text{error for observation $j$ in group $i$}), \tag 1 \\
& (\text{The numbers of observations in different groups may differ.}) \\[10pt]
\text{and } V_i = {} & \alpha + \beta x_i + (\text{error for observation $i$}). \tag 2
\end{align}
In the first case, let $\overline U_i$ be the sample average in group $i$. Then $U_{ij} - \overline U_i$ is the residual $\hat\varepsilon_{ij}$ and $\overline U_i - \overline U$ is the difference between the $i$th group mean and the grand mean. Then we have
\begin{align}
\sum_{ij} (U_{ij} - \overline U)^2 & = \text{total sum of squares} \\[10pt]
\sum_{ij} (\overline U_i - \overline U)^2 & = \text{explained part of the total sum of squares} \\[6pt]
\sum_{ij} (U_{ij} - \overline U_i) & = \text{unexplained part of the total sum of squares}
\end{align}
(Note that both indices $i,j$ are included in the first sum even though the index $j$ does not appear in the terms being added.)
In the second case, let $\widehat\alpha,\widehat\beta$ be the least-squares estimates of $\alpha,\beta$ and let $\overline V$ be the mean of all of $V_i.$ Then $V_i - (\widehat\alpha + \widehat\beta x_i)$ is the residual $\varepsilon_i.$ We have
\begin{align}
\sum_i (V_i - \overline V)^2 & = \text{total sum of squares} \\[10pt]
\sum_i (\widehat\alpha+\widehat\beta x_i - \overline V)^2 & = \text{explained part of the total sum of squares} \\[10pt]
\sum_i (V_i - (\widehat\alpha+ \widehat\beta x_i))^2 & = \text{unexplained part of the total sum of squares}
\end{align}
In the group-means case, the null hypothesis is that there are no difference in group means; in the fitting-a-line case the null hypothesis is that the slope $\beta$ is zero, so no differences in $V$-values result from changes in the $x$-value.
In both cases, there is a test statistic:
$$
F = \frac{(\text{explained part})/\text{df}}{(\text{unexplained part})/\text{df}},
$$
where $\text{“df''}$ is not the same in every case. In the group means case, the number of degrees of freedom in the numerator is $1$ less than the number of groups. In the denominator is is $1$ less than the number of members of a group, then summed over all groups. In the fitting-a-line case, the number of degrees of freedom in the numerator is $2$ and in the denominator is $2$ less than the number of observations.
If we assume i.i.d. mean-$0$ normally distributed errors, then $F$ has a standard central $F$-distribution in both cases, if the null hypothesis is true, and has a stochastically larger non-central $F$-distribution if the null hypothesis is false.
