Fundamental group of a wedge sum, in general (e.g. when van Kampen does not apply)

Under certain conditions they do actually mean the existence of neighborhoods, containing the common point $x_0$ of the wedge, which do retract on $x_0$.

With this condition we are able to choose suitable open path-connected sets $A_i$ to compute the desired fundamental group.

For instance if you take the $n$ wedge of circles $S^1$ at $x_0$ you can compute its fundamental group via Van-Kampen. What is important here is that we have open neighborhoods $N_i$ containing $x_0$ for each circle $S^1$ of the wedge (just imagine a little open arc on $S^1$ which contains the center $x_0$ of the wedge) which do actually retract on $x_0$ (just shrink from both sides until you reach the center $x_0$ of the wedge).

Now you can indeed define $A_i$ to be the wedge of all $N_j$ for $j \neq i$ and wedge $S^1$ i.e. $$A_i=\left(\bigvee_{i \neq j}N_j\right) \vee S^1.$$ If you now take two different $A_i$'s, i.e. $A_i$ and $A_j$ and look at the intersection $A_i \cap A_j$ you'll see that it's simply the wedge of all $n$ neighboorhoods $N_i$ of $x_0$ and since all of them retract to $x_0$ we get $$\pi_1\left( \bigvee_{i=1}^nN_i\right) \cong \pi_1(\{x_0\})=0.$$

By a similiar argument using your retracts you can compute $$\pi_1(A_i) \cong \pi_1(S^1) \cong \mathbb Z.$$

Now by applying Van Kampen you get that the fundamental group of your wedge sum is free abelian of rank $n$.

Suppose $X=A_0\cup A_1$ and we have $x_0\in A_0\cap A_1$. Now the Siefert-VanKampen Theorem states $$\pi_1(X,x_0)\cong\pi_1(A_0,x_0)\ast_{\pi_1(A_0,x_0)\cap\pi_1(A_1,x_0)}\pi_1(A_1,x_0)$$ is an amalgam provided several assumptions are satisfied:

1) We must have that $A_0$ and $A_1$ are open in $X$.

2) We must have that $A_0$ and $A_1$ and $A_0\cap A_1$ are path connected.

Notice that this specializes to something like the following:

Suppose $X=A_0\cup A_1$ where $x_0\in A_0\cap A_1$ and $A_0$, $A_1$, and $A_0\cap A_1$ are open, path-connected subspaces of $X$. If $A_0\cap A_1$ is simply connected, then $\pi_1(X,x_0)\cong\pi_1(A_0,x_0)\ast\pi_1(A_1,x_0)$.

Now condition (1) explains why we need to be careful about wedge products; the subspaces $A_0$ and $A_1$ must be open in $X$. Thus your desired result about fundamental groups of wedge products has the following formulation for a sufficient condition:

Let $X=\bigvee_{\alpha}A_\alpha$ be a finite wedge product of spaces $A_\alpha$ at $x_0$. If $x_0$ has a simply connected open neighborhood $U$ such that $U\cup A_\alpha$ is open for all $\alpha$ and $A_\alpha\cap U$ is path connected, simply connected and open for all $\alpha$, then $$\pi_1(\bigvee_\alpha A_\alpha,x_0)\cong\ast_\alpha\pi_1(A_\alpha,x_0)$$

This can be proved using induction and the stated version of the SVK Theorem above.

As for wedge-versus-vee, that seems to be an unfortunate product of historical stuff...