Regular representation and matrix coefficients One way to make sense of the regular representation $\rho_\text{reg}$ of $G$ is to consider it as the ring of complex valued functions $\{f:G\to\mathbb{C}\}$ with the action of $G$ given as $gf(x)=f(xg)$.
For a representation $\rho:G\to GL(V)$ we define the matrix coefficient as $f_{\phi,v}=\phi(\rho_gv)$ where $v\in V$ and $u\in V^*$ is a linear functional.
Question: I am asked to prove that every irreducible representation of $G$ is isomorphic to a subrepresentation of the regular representation using matrix coefficients.
I can see that the matrix coefficient $f_{\phi,u}$ is a function from $G\to\mathbb{C}$ and thus I get an embedding $V\otimes V^*\to \rho_\text{reg}$. How can I show that this a morphism of representations though? And afterwards I know that $V\otimes V^*\simeq \text{End}(V)\simeq M_n(\mathbb{C})$ but where can I go from there?
Edit: I know how to prove it using Wedderburn-Artin so it should not be used in this case
 A: Question: "I am asked to prove that every irreducible representation of G is isomorphic to a subrepresentation of the regular representation using matrix coefficients."
Answer: You find an elementary proof in Fulton-Harris Proposition 3.29. If $W_i$ are the finite dimensional irreducible $G$-modules they prove there is an isomorphism
$$ \mathbb{C}[G] \cong \oplus_i End(W_i).$$
I do not see any mention of the Wedderburn-Artin theorem in the proof.
To prove that a mapping of vector spaces is a map og $G$-modules you must prove it "commutes with the action of $G$": If $\psi: V^*\otimes_k V \rightarrow \rho_{reg}$ is a $k$-linear map you must prove that for any elements $g\in G, x \in V^*\otimes_k V$
it follows $\psi(gx)=g\psi(x)$.
Example: Examples of such maps are "natural maps" such as the isomorphism
$$f:Hom_k(E\otimes_k F, G) \cong Hom_k(E,Hom_k(F,G)).$$
If $E,F,G$ are $G$-modules it follows $f$ is an isomorphism of $G$-modules.
Here is an explicit non-trivial example:
How to construct an isomorphism $\wedge^2 V \xrightarrow{\sim} \wedge^2 V\otimes U^{\prime}$ as representations of $S_5$?
Note: You write
"For a representation ρ:G→GL(V) we define the matrix coefficient as fϕ,v=ϕ(ρgv) where v∈V and u∈V∗ is a linear functional."
Can you check this? What is $\rho_g$ and should $u=\phi$?
If you mean $f_{\phi,v}:=\phi(\rho(g)v)$ you get for any $g\in G$ an element
$$f_{\phi,v}(g):=\phi(\rho(g)v) \in k,$$
and this is a map
$$f_{\phi,v}: G \rightarrow k.$$
You get a map
$$\psi: V^*\otimes V \rightarrow Hom(G,k)$$
defined by
$$\psi(\phi \otimes v)(g):=f_{\phi,v}(g).$$
We get
$$\psi(h(\phi \otimes v))(g)=\psi((h\phi)\otimes (hv))(g)=$$
$$f_{h\phi, hv}(g)=\phi(h^{-1}ghv).$$
We moreover get
$$h\psi(\phi \otimes v)(g):=(hf_{\phi,v})(g)=f_{\phi,v}(gh)=\phi(ghv).$$
It seems the map $\psi$ I define above is not a map of $G$-modules. Your map may not be this one. You must specify your map - write down an explicit formula.
Note: If $T:=k$ is the trivial $G$-module there is a canonical map
$$tr: V^*\otimes_k V \rightarrow T$$
of $G$-modules:
$$tr(g(\phi \otimes v))=tr((g\phi)\otimes gv):=(g\phi)(gv):=\phi(v):=gtr(\phi \otimes v).$$
Hence $V^*\otimes_k V$ is not irreducible in general. Hence if you construct a map
$$\psi': V^*\otimes_k V \rightarrow Hom(G,k)$$
it may be $ker(\psi')\neq (0)$.
Note: There are many ways to embed an irreducible module into the group algebra: If $V$ is a finite dimensional irreducible $G$ module and if $0 \neq v\in V$ is any non-zero vector there is a canonical surjective map
$$\rho_v: \mathbb{C}[G] \rightarrow V \rightarrow 0$$
defined by $\rho_v(x):=xv$. We get a direct sum decomposition
$$\mathbb{C}[G] \cong W \oplus V$$
where $W:= ker(\rho_v)$ and this gives an embedding $V \subseteq \mathbb{C}[G]$ as a sub-$G$-module.
If $V_1,..,V_n$ are all finite dimensional irreducible representations of $G$,
you may using this method to realize these as sub-$G$-modules of $\mathbb{C}[G]$.  The decomposition of $\mathbb{C}[G]$ into irreducible $G$-modules gives the formula
$$I1.\text{   }\mathbb{C}[G] \cong \oplus_{i=1}^n V_i^{\oplus dim(V_i)}.$$
If $V_i:=k\{e(i)_1,..,e(i)_{n_i}\}$ you get embeddings
$$\rho_{e(i)_j}: V_i \rightarrow \mathbb{C}[G]$$
for all $j$. I guess this gives the formula $I1$.
There is as you noted for every $\phi \in V^*$ a map
$$f_{\phi}: V \rightarrow Hom(G,k)$$
defined by
$$f_{\phi}(v)(g):=\phi(gv).$$
It follows
$$f_{\phi}(hv)=hf_{\phi}(v)$$
and the map is $G$-linear. Since $V$ is irreducible it follows this map is injective.
