Your problem is local, so let me just work over one point, or in other words in a real vector space $V$ with a symmetric bilinear form $g$ (note: the term sesquilinear does not make sense in this case).
There are two ways to associate a complex vector space to $V$ that we have to discuss separately:
(1) Complexification: $V^\mathbb{C}:=V\otimes_\mathbb{R}\mathbb{C}$ becomes a complex vectorspace by defining
$$
\lambda \cdot (v\otimes \mu) = v \otimes(\lambda \mu) \quad \forall \lambda \in \mathbb{C}, v\otimes\mu \in V\otimes_\mathbb{R} C=V^\mathbb{C}
$$
For the dimension you have $\dim_\mathbb{C}V^\mathbb{C}=\dim_\mathbb{R}V$. Now you can extend $g$ to a bilinear form $G$ or to a sesquilinear form $H$ on $V^\mathbb{C}$. They are defined as follows:
$$G(v\otimes \lambda , w\otimes \mu)=\lambda \mu g(v,w) \quad \text{and} \quad H(v\otimes \lambda , w\otimes \mu)=\lambda \bar{\mu} g(v,w) $$
Note that $G$ will be symmetric again (i.e. $G(X,Y)=G(Y,X))$, whereas $H$ turns out to be hermitian (i.e. $H(X,Y)=\overline{H(Y,X)}$). In geometry one usually works with $H$, because it naturally provides us with a norm on $V^\mathbb{C}$.
Note that what we have done in the preceeding paragraph does not depend on a complex structure $J$, it always works.
(2) Complex Structure: Assume now that you have an (real!) endomorphism $J$ on V with $J^2=-I$. (In the vector space case this is called a linear complex structure and in in geometry usually an almost complex structure, while the term complex structure is reserved for those (global) endomorphisms which are integrable.)
Now you can turn $V$ itseld into a complex vector space that I will denote with $V_\mathbb{C}$. Complex multiplication is defined as
$$
\lambda \cdot v = \Re \lambda \cdot v + \Im\lambda \cdot Jv \quad \forall \lambda\in\mathbb{C}, v\in V_\mathbb{C}=V
$$
For the dimensions you have $\dim_\mathbb{C} V_\mathbb{C} = \dim_\mathbb{R}/2$.
Note that $g$ is still a function on $V_\mathbb{C}\otimes_\mathbb{C}V_\mathbb{C}$, but in general there is no reason to assume that it will be sesquilinear. This is really a stronger statement, because it involves the endomorphism $J$ and even $J-$invariance is usually not enough. However if $g$ happens to be $J$ invariant, then we can define a hermitian sesquilinear form $h$ on $V_\mathbb{C}$ in the way you have mentioned:
$$
h(v,w)=g(v,w)-ig(v,Jw)
$$
A counterexample:
In the second case I have claimed that we cannot expect $g$ to be sequilinear and indeed there is an easy counterexample: Take $V=\mathbb{R}^2$ with the usual euclidean inner product $g$ and let $J$ be the rotation by $\pi/2$. This is an isometry so $g$ is clearly $J-$invariant. This $J$ induces a complex multiplication on $V$ that is identical to the usual one for the complex numbers, so $V_\mathbb{C}=\mathbb{C}$.
Let $v=(1,0)$, then $i\cdot v = (0,1)$, but
$$
g(i \cdot v,v) = 0 \neq i = i g(v,v),
$$
hence $g$ is not sesquilinear.
