We start with the population model:
$$\begin{aligned}y &=E(y|{\bf x)}+e \\ E(e|{\bf x}) &=0 \end{aligned}$$
where $m({\bf x})=E(y|x)$ is the conditional expectation function (CEF) and e is the CEF error. It is important to note that the equations above are a definition. Here $E(e|{\bf x})=0$ is a defintion, not a restriction.
It can be shown that $m({\bf x})$ is the Best Predictor of $y$ in the sense that it minimizes the mean squared prediction error $E\left[(y-g({\bf x}))^2\right]$. If we could obtain $m({\bf x})$ then we would have the best overall predictor of $y$.
However, we do not know the functional form of $m({\bf x})$. We could assume it is linear, but this might be overly-restrictive. Why would it be linear?
Instead we consider the Best Linear Predictor of $y$. A linear predictor of $y$ is function of the form ${\bf x}'\boldsymbol\beta$. The best linear predictor minimizes
$$E\left[ (y- {\bf x}'\boldsymbol\beta)^2\right] \tag{1}$$
If the variance matrix of $\bf x$, $E({\bf xx}')$, is positive definite (i.e, invertible, i.e., nonsingular) then we can find a unique solution for $\beta$ in (1) by taking and solving the first order condition. We get
$$\boldsymbol\beta=\left(E[{\bf xx}'] \right)^{-1}E[{\bf x}y] \tag{2}$$
By pluggin (2) into ${\bf x}'\boldsymbol\beta$ we get the best linear predictor also called the Linear Projection of $y$ on $\bf x$
$$\begin{aligned}L(y|{\bf x})&={\bf x}'\boldsymbol\beta \\ \text{where} \ \ \ \ \boldsymbol\beta &=\left(E[{\bf xx}'] \right)^{-1}E[{\bf x}y]\end{aligned}$$
The Projection Error is then:
$$e=y-{\bf x}'\boldsymbol\beta \tag{3}$$
and we can see that $E({\bf x}e)=0$
$$\begin{aligned} E[{\bf x}e] &= E[{\bf x}(y-{\bf x}'\boldsymbol\beta)] \\ &= E[{\bf x}y]-E[{\bf xx}'] \left(E[{\bf xx}']\right)^{-1}E[{\bf x}y] \\ &={\bf 0} \end{aligned}$$
Since $\bf x$ has a constant we get $E(e)=0$ as well.
Now lets take (3) rearrange it to have y on the left side. Then separate the constant out of $\bf x$ and take expectations on both sides
$$E[y] =E[{\bf x}'\boldsymbol\beta]+E[\beta_0]+E[e] $$
Note that $E[\beta_0]=\beta_0$ and $E[e]=0$. Then solve for $\beta_0$ to get same as Wooldridge.
$$\beta_0=E[y]-E[{\bf x}]'\boldsymbol\beta \tag{4}$$
Now subtract (4) from (3) to get,
$$y-E[y]=({\bf x}-E[{\bf x}])'\boldsymbol\beta + e \tag{5}$$
Because $({\bf x}-E[{\bf x}])$ and $e$ are uncorrelated, (5) is also a linear projection and we can find $\boldsymbol\beta$
$$\begin{aligned}\boldsymbol\beta & =\left(E\left[ ({\bf x}-E[{\bf x}])({\bf x}-E[{\bf x}])'\right]\right)^{-1} E\left[ (y-E[y])(y-E[y])'\right] \\ &= [\text{Var}({\bf x})]^{-1} \text{Cov}({\bf x}y)\end{aligned}$$.
Thus we have our Linear Projection Model
$$\begin{aligned} & y=\beta_0 +{\bf x}'\boldsymbol\beta +e \\ \text{where} & \\ & L(y|1,{\bf x})= \beta_0 + {\bf x}'\boldsymbol\beta \\ & \boldsymbol\beta=[\text{Var}({\bf x})]^{-1} \text{Cov}({\bf x}y) \\ & \beta_0=E[y]-E[{\bf x}]'\boldsymbol\beta \end{aligned}$$