Explicit relation between dual and adjoint of a linear map Let $V$ and $W$ be finite-dimensional vector spaces over some arbitrary field $K$, and let $V^\ast$ and $W^\ast$ be their respective dual spaces. Let $f:V \rightarrow W$ be a linear map.
a) Define the dual of $f$ as the map $f^\ast : W^\ast \rightarrow V^\ast$, $e \mapsto e \circ f$.
b) Suppose we define two fixed non-degenerate bilinear forms $\langle \cdot,\cdot \rangle_V : V^2 \rightarrow K$, $\langle \cdot,\cdot \rangle_W : W^2 \rightarrow K$. Define the adjoint of $f$ as the map $\bar{f}: W \rightarrow V$ satisfying
$\langle v,\bar{f}(w) \rangle_V = \langle f(v),w \rangle_W$.
Since the bilinear forms are non-degenerate, the linear maps $\phi_V : V \rightarrow V^\ast$ and $\phi_W : W \rightarrow W^\ast$ given by
$$ \phi_V(v)(v_0) = \langle v,v_0 \rangle_V, \qquad \phi_W(w)(w_0) = \langle w,w_0 \rangle_W $$
are isomorphisms. Given this, how can we express $\bar{f}$ explicitly in terms of $f^\ast$ (and if required, $\phi_V$ and $\phi_W$)? Alternatively, how can we prove that the map $\bar{f}$ is guaranteed to exist?
EDIT: As pointed out by @levap I have fixed the definition of $\phi_W$.

Here's an approach using matrices.
By a choice of basis identify $V=K^n$ and $W=K^m$, elements are column vectors. We can identify their duals $V^*$ and $W^*$ with spaces of row vectors so that the usual matrix product serves as application of linear forms to vectors. In this setting, when $f\colon V\to W$ is given by an $m\times n$-matrix $A$ such that $f(v) = Av$, its dual $f^*\colon W^*\to V^*$ is given by $f^*(\beta) = \beta A$ since then
$$
(f^*(\beta))(v) = \beta A v = \beta(f(v)).
$$
Your bilinear forms are given by square matrices, namely
$$
\langle v, v'\rangle_V = v^t X v' \qquad\text{and}\qquad
\langle w, w'\rangle_W = w^t Y w'.
$$
for invertible matrices $X_{ij}=(\langle e_i, e_j\rangle_V)_{i,j=1,\dots,n}$ and $Y_{ij}=(\langle e_i, e_j\rangle_W)_{i,j=1,\dots,m}$.
Your isomorphisms $\phi_V \colon V\to V^*$ and $\phi_W\colon W\to W^*$ are given by $\phi_V(v)(v') = v^t X v'$ and $\phi_W(w)(w') = w^t Y w'$.
Now let's find the $n\times m$-matrix $B$ describing the adjoint $\bar f\colon W\to V$ as $\bar f(w) = Bw$. We want to have
$$
\langle v, \bar f(w) \rangle_V = \langle f(v), w\rangle_W,
$$
which can be written as
$$
v^t X B w = (Av)^t Y w = v^t A^t Y w.
$$
So we have $XB=A^t Y$ and hence $B = X^{-1} A^t Y$.
