Such a map $\Phi$ does not exist for $n=5$ and $k=2$.

Let me in fact show a stronger claim: There exists no $\mathbb{R}$-linear map $\Phi : \mathbb{R}^5 \to \mathbb{C}^{2\times 2}$ such that every
five reals $x_{1},x_{2},\ldots,x_{5}\in\mathbb{R}$ satisfy
\begin{equation}
\det\left( \Phi\left( x_{1},x_{2},\ldots,x_{5}\right)
\right) =e_{2}\left( x_{1},x_{2},\ldots,x_{5}\right)
.
\label{eq.darij1.1}
\tag{1}
\end{equation}

*Proof.* Assume the contrary. Thus, such a map $\Phi$ exists. Consider it. Write it as
\begin{align}
\Phi\left( x_{1},x_{2},\ldots,x_{5}\right)
= \begin{pmatrix}
f_{11}\left( x\right) & f_{12}\left( x\right) \\
f_{21}\left( x\right) & f_{22}\left( x\right)
\end{pmatrix}
,
\label{eq.darij1.2}
\tag{2}
\end{align}
where the $f_{ij}\left( x\right) $ are four linear forms in $x_{1},x_{2},\ldots,x_{5}$ with complex coefficients (that is, four linear maps from $\mathbb{R}^5$ to $\mathbb{C}$).
Thus, by our assumption, any five reals $x_{1},x_{2},\ldots,x_{5}$ satisfy
\begin{align}
& e_{2}\left( x_{1},x_{2},\ldots,x_{5}\right) \nonumber\\
& =\det\left( \Phi\left( x_{1},x_{2},\ldots
,x_{5}\right) \right) \qquad\left( \text{by \eqref{eq.darij1.1}}
\right) \nonumber\\
& =\det
\begin{pmatrix}
f_{11}\left( x\right) & f_{12}\left( x\right) \\
f_{21}\left( x\right) & f_{22}\left( x\right)
\end{pmatrix}
\qquad\left( \text{by \eqref{eq.darij1.2}}\right)
.
\label{eq.darij1.3}
\tag{3}
\end{align}
Let us extend the four linear forms $f_{ij}$ from $\mathbb{R}^5$ to $\mathbb{C}^5$ in the obvious way (i.e., preserving their coefficients). Thus, the $f_{ij}$ are now four $\mathbb{C}$-linear forms from $\mathbb{C}^5$ to $\mathbb{C}$.

Then, \eqref{eq.darij1.3} is an equality between two polynomials in $x_{1},x_{2},\ldots,x_{5}$.
Thus, since we know that it holds for all reals $x_{1},x_{2},\ldots,x_{5}$, we
conclude that it holds for all complex numbers $x_{1},x_{2},\ldots,x_{5}$.

Now, recall that the $f_{ij}$ are linear forms on $\mathbb{C}^{5}$. Let $W$ be
the $\mathbb{C}$-vector subspace $\operatorname*{Ker}\left( f_{21}\right)
\cap\operatorname*{Ker}\left( f_{22}\right) $ of $\mathbb{C}^{5}$. The
dimension of this subspace $W$ is at least $3$ (since it is defined by two
linear equations in $\mathbb{C}^{5}$), thus larger than $5/2$.

But $e_{2}\left( x_{1},x_{2},\ldots,x_{5}\right) =x_{1}x_{2}+x_{1}
x_{3}+\cdots+x_{4}x_{5}$ is a symmetric bilinear form on $\mathbb{C}^{5}$.
This form is easily seen to be nondegenerate, since it is represented by the
invertible symmetric matrix $\dfrac{1}{2}
\begin{pmatrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & 1 & 1 & 1\\
1 & 1 & 0 & 1 & 1\\
1 & 1 & 1 & 0 & 1\\
1 & 1 & 1 & 1 & 0
\end{pmatrix}$. Hence, a known fact (see, e.g.,
https://math.stackexchange.com/questions/4303680/ ) says that it cannot become
identically zero when restricted to a subspace of $\mathbb{C}^{5}$ whose
dimension is larger than $5/2$. In particular, it cannot become identically
zero when restricted to $W$ (since $W$ has dimension larger than $5/2$).
Hence, the corresponding quadratic form also cannot become identically zero
when restricted to $W$ (because if a symmetric bilinear form is nonzero, then
the the corresponding quadratic form is also nonzero). In other words, there
exists some $x=\left( x_{1},x_{2},\ldots,x_{5}\right) \in W$ such that
$e_{2}\left( x_{1},x_{2},\ldots,x_{5}\right) \neq0$. However, $x\in W$
entails $f_{21}\left( x\right) =0$ and $f_{22}\left( x\right) =0$ and
thus
\begin{align*}
\det
\begin{pmatrix}
f_{11}\left( x\right) & f_{12}\left( x\right) \\
f_{21}\left( x\right) & f_{22}\left( x\right)
\end{pmatrix}
=\det
\begin{pmatrix}
f_{11}\left( x\right) & f_{12}\left( x\right) \\
0 & 0
\end{pmatrix}
=0,
\end{align*}
so that \eqref{eq.darij1.3} rewrites as $e_{2}\left( x_{1},x_{2},\ldots
,x_{5}\right) =0$. This contradicts $e_{2}\left( x_{1},x_{2},\ldots
,x_{5}\right) \neq0$. This contradiction completes our proof. $\blacksquare$

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