Here is the explanation of how the ANOVA works, based of off Bickel and Doksum's Mathematical Statistics (1977), Chapter 7. This is a geometric view (read as Linear Algebraic view) and hopefully can provide some intuition on what exactly the F-test and ANOVA is doing. Some prerequisite understanding required is the concept of linear subspaces, image spaces, and projections, orthogonality of subspaces, and norms of vectors.
Continuous Model
Your continuous model has the form
$Y_i = \beta_0 + \beta_1 X_{1,i} + \epsilon_i$
where $X_{1,i}$ is the i-th observation of temperature. Your design matrix looks like:
$X_{cont} = \begin{bmatrix} 1 & x_{1,1} \\ 1 & x_{1,2} \\ \vdots & \vdots \\ 1 & x_{1,n}\end{bmatrix}$ where your $x_{1,i}$ for $i = 1,...,n$ are your temperature observations.
Categorical Case
In the categorical case, your design matrix will look different. You model will now be of the form:
$Y_i = \beta_0 + \beta_1T_{2,i} + \beta_2 T_{3,i} + ... + \beta_{n-1} T_{n,i} + \epsilon_i$. Note that $T_{1,i}$ is not in the model, as it has been pooled into the $\beta_0$ term. We cannot have $\beta_0$ exist along with coefficient for all levels of your categorical temperature for model identifiable issues assuming OLS where $\epsilon_i \sim N(0,\sigma^2)$.
So your design matrix will look the following, in the case of your data set with 5 distinct categories :
$X_{cat} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1\end{bmatrix}$ and so on. Of course this means your estimate of $E(Y|T=T_1)$ is $\beta_0$ and your estimate for $E(Y|T=T_i)$ where $i > 1$ is $\beta_0 + \beta_{i-1}$.
Connecting the two cases
Note that in your case with 5 categories, $X_{cont} = \begin{bmatrix} 1 & x_{1,1} \\ 1 & x_{1,2} \\ 1 & x_{1,3} \\ 1 & x_{1,4} \\ 1 & x_{1,5}\end{bmatrix}$, where $x_{1,i}$ is assumed to be any real number (of course, this is an assumption that may not always hold depending how extreme your data is, but in general, is the assumption for OLS). The column span (image space) of $X_{cont}$ lies in the image space of $X_{cat}$. To see this: consider the the image space of $X_{cat}$ and $X_{cont}$.
For the image space of the design matrix in the categorical case:
$\omega := Img(X_{cat}) = \lbrace{\mathbf{v} \in R^5 : \alpha_1 \mathbf{1} + \alpha_2 \mathbf{e}_2 + \alpha_3 \mathbf{e}_3 + \alpha_4 \mathbf{e}_4 + \alpha_5 \mathbf{e}_5\rbrace}$, where $\mathbf{1}$ is a 1 x 5 vector of ones, and $e_i$ is the i-th standard basis vector; and $\alpha_j$ can take on any real number. In short, these are just the columns of $X_{cat}$ put into vector form, and we are just taking linear combinations of these vectors to get the image space.
For the image space of the design matrix in the continuous case,
$\omega_0 := Img(X_{cont}) = \lbrace{\gamma_1 \mathbf{1} + \gamma_2 \mathbf{x} \rbrace}$ where $\mathbf{x}$ is the 1 x 5 column vector of temperature observations and $\mathbf{x}$ looks like this $\begin{bmatrix}x_{1,1} \\ x_{1,2} \\ x_{1,3}\\ x_{1,4} \\ x_{1,5}\end{bmatrix}$. Again, this is just taking the columns of $X_{cont}$ as vectors, and taking linear combinations of such vectors.
If I look at image space of $X_{cat}$, and I set $\alpha_1 = x_{1,1}$ and $\alpha_2 = x_{1,2}-x_{1,1}$ and so on, so that we end with $\alpha_5 = x_{1,5}-x_{1,1}$, then we get a vector $\begin{bmatrix}x_{1,1} \\ x_{1,2} \\ x_{1,3}\\ x_{1,4} \\ x_{1,5}\end{bmatrix}$. So clearly this vector lies in the image space of $X_{cat}$.
Now if I look at the image space of $X_{cont}$, and I set $\gamma_1 = 0$ and $\gamma_1 = 1$, then I also get the same vector $\begin{bmatrix}x_{1,1} \\ x_{1,2} \\ x_{1,3}\\ x_{1,4} \\ x_{1,5}\end{bmatrix}$. In fact, any vector in the image space of $X_{cont}$ is in the image space of $X_{cat}$. Thus we say the continuous model is "nested" in the categorical model, in the sense that the $\omega$ and $\omega_0$ have a nested relationship.
Technical clarification: (in order to call an image space a "subspace", we must allow the image space to contain the 0, and be closed under vector addition and scalar multiplication. Thus we must place conditions on $\alpha_i$ and $\gamma_i$ to meet these requirements of a subspace. Here is not an issue as we can simply define the $\alpha_i$ and $\gamma_i$ as elements of R).
ANOVA
ANOVA is defined by Bickel and Docksum in terms of projections. I do not have the book on hand right now, but I will update this answer to provide page numbers.
The ANOVA produces a test statistic given by $\frac{||P_\omega Y - P_{\omega_0}Y||^2/(\dim \omega - \dim \omega_0)}{||Y-P_{\omega} Y||^2/(n-\dim\omega)}$. Here $P_{\omega}$ is the projection of your $Y$ responses onto the $\omega$ linear space, and likewise for $P_{\omega_0}$. But what is happening here?
First note $P_{\omega}Y$ essentially represents your regression estimates in the categorical model. Then $P_{\omega_0}Y$ represents your regression estimates in the continuous model. Also note that $Y-P_{\omega_0}Y$ is the vector of residual of the continuous model and $Y-P_{\omega}Y$ is the vector of residuals in the categorical model. Of course we can square them to get sum of square errors. Now we claim that $||Y-P_{\omega_0}Y||^2 = ||Y-P_{\omega}Y||^2 + ||P_\omega Y - P_{\omega_0}Y||^2$. You can easily show this by noting that $Y-P_\omega Y$ is orthogonal to any vector in $\omega$ and that since $P_\omega Y - P_{\omega_0}Y$ must also lie in $\omega$, it is also orthogonal to $Y-P_\omega Y$, so it goes like this:
\begin{align*}
Y-P_{\omega_0}Y &= Y- P_{\omega}Y + P_{\omega}Y - P_{\omega_0}Y \\
||Y-P_{\omega_0}Y||^2 &= ||Y- P_{\omega}Y + P_{\omega}Y - P_{\omega_0}Y||^2 \\
&= ||Y- P_{\omega}Y||^2 + ||P_{\omega}Y - P_{\omega_0}Y||^2 && \text{Pythagorean Theorem}
\end{align*}
So some algebra will show us that $||P_{\omega}Y - P_{\omega_0}Y||^2 = ||Y-P_{\omega_0}Y||^2 - ||Y- P_{\omega}Y||^2$ which is the difference in residuals between the two models. This makes sense since $||Y-P_{\omega_0}||^2$ is expected to be larger than a more refined model with SSE given by $||Y-P_{\omega}||^2$.
So now we know that the ANOVA where $F = \frac{||P_\omega Y - P_{\omega_0}Y||^2/(\dim \omega - \dim \omega_0)}{||Y-P_{\omega} Y||^2/(n-\dim\omega)}$ is really $\frac{(||Y-P_{\omega_0}Y||^2 - ||Y- P_{\omega}Y||^2)/(\dim \omega - \dim \omega_0)}{||Y-P_{\omega} Y||^2/(n-\dim\omega)}$. (We divide by some factors to deal with MSE rather than SSE as a way to standardize and also as a way to make the statistical distribution into an F distribution, for reasons that should be for another topic.)
Conclusion and TLDR
So now we see that the ANOVA is really just comparing how much better a more refined model (given by subspace $\omega$) will perform, in terms of residuals, when compared to a less refined model (given by subspace $\omega_0$). And "model 2 is more refined than model 1" is the same as saying the image space of model 2 contains the image space of model 1, or $\omega_0 \subseteq \omega$. The nesting requirement is critical so that we may invoke orthogonality and pythagorean theorem when decomposing $||Y-P_{\omega_0}Y||^2$.
In the context of your data set
Your output shows a p-value of 0.11. This fails to reject the null hypothesis at the 0.05 level. Well what is the null?
Note that in the above exposition, we were essentially testing if your regression model, which estimates $E(Y|X)$, lies in the $\omega_0$ subspace or in the $\omega$ subspace, so we are really testing if $H_0: E(Y|X) \in \omega_0$ versus the alternate $H_a: E(Y|X) \in \omega$. Since we fail to reject $H_a$, we fall back to our null, which is $E(Y|X) \in \omega_0$, that is, the categorical model does not provide a significant reduction in SSE to warrant its use over the continuous model.
Rejecting the null means that the difference between the SSE's the is, the difference between $||Y-P_{\omega_0}Y||^2$ and $||Y-P_{\omega}Y||^2$ is large enough for one to be considered "better".
tempto the regressor matrixfactor(temp)and find out whether the rank of the regressor matrix remains the same or not. $\endgroup$