DaftWulie's answer to Extending a square matrix to a Unitary matrix says that extending a matrix into a unitary cannot be done unless there's constraints on the matrix. What are the constraints?

A necessary and sufficient condition is that, given an $ntimes n$ matrix $M$, you can construct a $2ntimes 2n$ unitary matrix $U$ provided the singular values of $M$ are all upper bounded by 1.

Sufficiency

To see this, express the singular value decomposition of $M$ as
$$
M=RDV
$$
where $D$ is diagonal and $R$, $V$ are unitary. Now define
$$
U=left(begin{array}{cc}
M & Rsqrt{mathbb{I}-D^2}V \
Rsqrt{mathbb{I}-D^2}V & -M
end{array}right),
$$
which we can only do if the singular values are no larger than 1. Let's verify that it's unitary
begin{align*}
UU^dagger&=left(begin{array}{cc}
RDV & Rsqrt{mathbb{I}-D^2}V \
Rsqrt{mathbb{I}-D^2}V & -RDV
end{array}right)left(begin{array}{cc}
V^dagger DR^dagger & V^daggersqrt{mathbb{I}-D^2}R^dagger \
V^daggersqrt{mathbb{I}-D^2}R^dagger & -V^dagger DR^dagger
end{array}right) \
&=left(begin{array}{cc}
RD^2R^dagger+R(mathbb{I}-D^2)R^dagger & 0 \
0 & RD^2R^dagger+R(mathbb{I}-D^2)R^dagger
end{array}right) \
&=mathbb{I}.
end{align*}

Necessity

Imagine I have a matrix $M$ with a singular value $lambda>1$ and corresponding normalised vector $|lambdarangle$. Assume I construct a unitary
$$
U=left(begin{array}{cc} M & A \ B & C end{array}right).
$$
Let's act $U$ on the state $left(begin{array}{c} |lambdarangle \ 0 end{array}right)$. We get
$$
Uleft(begin{array}{c} |lambdarangle \ 0 end{array}right)=left(begin{array}{c} M|lambdarangle \ B|lambdarangle end{array}right).
$$
This output state must have a norm that is at least the norm of $M|lambdarangle$, i.e. $lambda>1$. But if $U$ is a unitary, the norm must be 1. So it must be impossible to perform such a construction if there exists a singular value $lambda>1$.

$newcommand{bs}[1]{boldsymbol{#1}}$Here is a slightly different way to prove what the other excellent answer did.

Note that a matrix $U$ is unitary if and only if it sends orthonormal bases into orthonormal bases.
This, in particular, means that if $U$ is unitary then $|Ubs v|=1$ for any $bs v$ with $|bs v|=1$.

Let us write the SVD of $M$ as $Mbs u_k=s_kbs v_k$, where $s_kge0$ are the singular values of $M$.

Note that if $U$ is an extension of $M$, then $Ubs u_k=s_k bs v_k+bs w_k$ for some $bs w_k$ orthogonal to $bs v_k$ (and more generally to the whole range of $M$).

If follows that if, for any $k$, $s_k>1$, then $|Ubs u_k|>1$, and thus $U$ is not unitary.

On the other hand, if $s_kle1$ for all $k$, let us show how can always construct a unitary $U$ that contains $M$ as a submatrix.
Let us denote with $bs voplus bs 0$ the vectors in the extended $2n$-dimensional space that are built by appending zeros to the $n$-dimensional vector $bs v$, and with $bs 0oplusbs v$ the vectors that are equal to $bs v$ in the last $n$ dimensions by zero in the first $n$ ones.
Being ${bs u_k}_k$ a basis for the original space, it follows that ${bs u_koplus bs 0,bs0oplusbs u_k}_k$ is a basis for the extended space.

We will define $U$ through its action on the vectors $u_koplus bs 0$ and $bs0oplus u_k$ as follows:
begin{align}
U(bs u_koplus bs 0)&=s_k(bs v_koplusbs 0)+sqrt{1-s_k^2}(bs 0oplus bs v_k) \
U(bs0 oplus bs u_k)&=sqrt{1-s_k^2}(bs v_koplusbs 0)-s_k(bs 0oplus bs v_k).
end{align}

One can then check that all of these output vectors form an orthonormal system in the extended space, and thus $U$ is unitary.