In this case, it is well-known that the likelihood-ratio test statistic for testing $\eqref{0}$ is given by
\begin{align*}
\Lambda = n\log\left(\frac{RSS(M_0)}{RSS(M)}\right), \tag{1}\label{1}
\end{align*}
and the F-test statistic for testing $\eqref{0}$ is given by
\begin{align*}
F = \frac{\frac{RSS(M_0) - RSS(M)}{p - 1}}{\frac{RSS(M)}{n - p - 1}}. \tag{2}\label{2}
\end{align*}
Under the normality assumption, we know that $F$ in $\eqref{2}$ has an exact $F_{p - 1, n - p - 1}$ distribution (check this answer for the proof). But does $\Lambda$ in $\eqref{1}$ have an exact $\chi_{p - 1}^2$ distribution? To the best of my knowledge, no. It is just asymptoticallyNO! It is just asymptotically $\chi_{p - 1}^2$ distributed $\chi_{p - 1}^2$ distributed. This asymptotic result is a consequence of the famous Wilk's theorem. At the end of this answer, I derived the exact distribution of $\Lambda$, from which we can also see the asymptotic distribution of $\Lambda$ is $\chi_{p - 1}^2$.
It can be seen that the two tests even gave contradictory results at the significance level $\alpha = 0.05$. Clearly, the $F$-test did a better job (as expected), and this empirical result supports my theoretical "Case 1" analysis conclusion -- even under the much stronger (and usually unrealistic) normality assumption, the $F$-test and the likelihood-ratio test still do not concur! Therefore, the first sentence of your post "the t-test and F-test we use to get the significance of our regression results are derived from the likelihood ratio test" is a misconception -- although they are numerically related (e.g., if $\Lambda$ in $\eqref{1}$ is large, then $F$ in $\eqref{2}$ is large too because $\eqref{2}$ is a monotonic function of $\eqref{1}$), it is by no means to say that (the null distribution of) the $F$-test is "derived" from the likelihood-ratio test!
Exact distribution of $\Lambda$ in Case 1.
It follows by $\eqref{1}$ and $\eqref{2}$ that
\begin{align*}
F = \frac{n - p - 1}{p - 1}(e^{\Lambda/n} - 1), \tag{3}\label{3}
\end{align*}
whence the Jacobian of transformation $\eqref{3}$ is
\begin{align*}
J = \frac{n - p - 1}{n(p - 1)}e^{\lambda/n}. \tag{4}\label{4}
\end{align*}
Since the pdf of a $F_{p - 1, n - p - 1}$ r.v. is
\begin{align*}
f_F(x) = \frac{1}{B\left(\frac{p - 1}{2}, \frac{n - p - 1}{2}\right)}
\left(\frac{p - 1}{n - p - 1}\right)^{\frac{p - 1}{2}}x^{\frac{p - 1}{2} - 1}
\left(1 + \frac{p - 1}{n - p - 1}x\right)^{-\frac{n}{2} + 1}, \tag{5}\label{5}
\end{align*}
it follows by $\eqref{3}, \eqref{4}$ and $\eqref{5}$ that the pdf of $\Lambda$ is given by
\begin{align*}
f_\Lambda(\lambda) = \frac{1}{B\left(\frac{p - 1}{2}, \frac{n - p - 1}{2}\right)}
e^{-\lambda/2}
\left(e^{\lambda/n} - 1\right)^{\frac{p - 1}{2} - 1}\frac{1}{n}e^{\frac{2\lambda}{n}}. \tag{6}\label{6}
\end{align*}
Evidently, $\eqref{6}$ is different from the pdf of a $\chi_{p - 1}^2$ r.v., whose pdf is
\begin{align*}
g(y) = \frac{1}{2^{\frac{p - 1}{2}}\Gamma\left(\frac{p - 1}{2}\right)}y^{\frac{p - 1}{2} - 1}e^{-y/2}. \tag{7}\label{7}
\end{align*}
On the other hand, as $n \to \infty$, $f_\Lambda(y)$ does converge to $g(y)$ point-wisely (the verification is not too hard but tedious, where we need to use $e^x \sim 1 + x$ as $x \to 0$ and $\Gamma(z) \sim \sqrt{2\pi}z^{z - 1/2}e^{-z}$ as $z \to \infty$), showing that $\Lambda \to_d \chi_{p - 1}^2$ as $n \to \infty$, as Wilk's theorem asserted.
