Yhyjyk

Let $G$ and $H$ be free groups and $g in G$. Is there a difference between $langle H, Grangle$ and the free product $H*G$?. In particular is $langle H ,g rangle = H * langle g rangle$?

In general, yes there is a difference between these two groups. For instance, consider the subgroups $2mathbf Z$ and $3mathbf Z$. Then we have $langle 2mathbf Z , 3mathbf Zrangle = mathbf Z$, but $2mathbf Zast3mathbf Z = 2mathbf Zastlangle 3ranglecongmathbf Zastmathbf Z$ is not an abelian group, since we do not impose any commutation relations between elements of (the isomorphic copy of) $2mathbf Z$ and the elements of (the isomorphic copy of) $3mathbf Z$.

As @Dietrich Burde suggests you can take $H=G=$(any free group $F_n,nge1$) and then $langle G,Hrangle=H=G$ and $H*G$ is not isomorphic to $H$

Since none of these points are in the answers above, although they answer your question, I'd like to add some comments. I won't add a new counterexample, since the other answers already cover that, just add a more abstract viewpoint.

First of all, there is an essential difference between $langle G,Hrangle$ and $G*H$, namely $langle G,Hrangle$ presupposes that $G$ and $H$ are embedded in some larger group, which I'll call $K$, whereas the free product does not. The group operation on $K$ may impose relations on $G$ and $H$ that are not present in the free product (which has no relations between the elements of $G$ and the elements of $H$).

What this means concretely is that if we consider the homomorphism $phi:G*Hto K$ given by the universal property of the free product, its image is $langle G,Hrangle$, and its kernel tells us whether or not $G*H$ and $langle G,Hrangle$ are isomorphic (technically this only tells us if they're naturally isomorphic, but let's ignore that tangent).

The elements of the kernel of $phi$ will be the relations between elements of $G$ and $H$ imposed by the group operation on $K$.

If we look at giannispapav/Dietrich Burde's counterexamples, if we let $K=G=H=F_n$ be free groups, then $langle G,Hrangle =K$, and $G*H=F_2n$, and the map sends the word $g_1h_1g_2h_2cdots g_nh_n$ in $G*H$ to the result of evaluating this product in $K$.

The relations are $g_Gg^-1_H=1$, where $g_G$ is an element $g$ of $K$ regarded as an element of $G$, $g^-1_H$ is the inverse of the same element $g$ of $K$ regarded as an element of $H$.

The response is there is no difference if one of them is a group with at least one has rank above one, because taking their presentations
$G=langle g_1,g_2,...|quad rangle$ and $H=langle h_1,h_2,...|quad rangle$ then you get for the free product the presentation
$$G*H=langle g_1,g_2,...,h_1,h_2,...|quad rangle.$$
The particular case $H*langle grangle=langle h_1,h_2,...,g|quad rangle$ follows.