I'm a theoretical physicist working on general relativity, I am familiar with differential geometry but there's something I've never understood. Let there be an embedding
$$\phi:S\rightarrow M$$
where $M$ is a $d$ dimensional manifold and $S$ is a codimension-k surface. The way to define the normal 1-forms to $S$ is (I think?) to find the kernel of the pullback $\phi^*:T^*M\rightarrow T^*S$. That is, to find all 1-forms $\tilde{n}^a$ with $a=1,\dots,k$ such that
$$\phi^*\tilde{n}^a=0$$
Let's say we use coordinates $\sigma^A$ with $A=1,\dots,d-k$ to parametrize $S$ and we parametrize $M$ with coordinates $y^a=(c^i,x^A)$ where $a=1,\dots,d$, $i=1,\dots,k$, $A=1,\dots,d-k$ and $\phi^* dx^A=d\sigma^A$. The idea is that $c^i$ represents the implicit equations that determine the embedding of $S$ in $M$. Now, a 1-form in $T^*M$ can be written as
$$\tilde{\rho}=\rho_i \tilde{dc}^i +\rho_A dx^A$$
so its pullback is
$$\phi^* \tilde{\rho}=\rho_i(\sigma) \frac{\partial c^i}{\partial \sigma^A}+\rho_A(\sigma)\tilde{d\sigma}^A$$
We get that we can define $k$ 1-forms
$$\tilde{n}^i= \big(\tilde{dc}^i-a^i_A \tilde{dx}^A)$$
where $a^i_A$ is not fully determined but its pullback must be
$$\phi^* a^i _A =\frac{\partial c^i}{\partial \sigma^A}$$
Does this make any sense? Why is the space of normal 1-forms not fully determined? Would adding a metric tensor to $M$ fully determine the normal to the embedding?
Unless I’m misunderstanding what you’re actually trying to get at, the answer seems pretty simple, yes it’s not uniquely determined. In my answer, I will avoid the confusing term ‘normal 1-form’.
First of all, talking about the map $\phi$ is just extra distraction. We have an embedding, so we may as well replace systematically $S$ with $\phi[S]$ in all the discussion. So, suppose $S\subset M$ is an embedded submanifold. You are now trying to find all 1-forms on $M$ whose pullback to $S$ by the inclusion map is $0$, i.e essentially all 1-forms on $M$ whose ‘restriction’ to $S$ is $0$. Next, even the manifolds are a distraction; by focusing your attention one tangent space at a time, this is really a question of linear algebra, so that is what I’ll focus on henceforth.
So, suppose I have a $n$-dimensional vector space $V$ over some field $\Bbb{F}$ and a $k$-dimensional subspace $W$ of $V$. You are now trying to find all linear maps $\lambda:V\to\Bbb{F}$ such that the restriction $\lambda|_W:W\to\Bbb{F}$ is the zero linear map. This is a very well-known space, it’s called the annihilator of $W$ in $V$, and is denoted as $W^0$. So, to be super explicit, by definition, \begin{align} W^0=\{\lambda\in V^*\,:\lambda|_W=0\}. \end{align} You can easily check that $W^0$ is a subspace of $V^*$. The space $W^0$ is measuring the non-uniqueness of linear functionals on $V$ when restricting them to $W$. More explicitly, for two linear maps $\lambda_1,\lambda_2:V\to\Bbb{F}$, we have that $\lambda_1|_W=\lambda_2|_W$ if and only if their difference $\lambda=\lambda_1-\lambda_2$ vanishes on $W$, i.e if and only if $\lambda_1-\lambda_2\in W^0$. Hence, in the case that $W^0$ is positive-dimensional, just because two linear functionals on $V$ agree on a subspace $W$, it does not imply they are equal on all of $V$. Another way to say this is that if you start with a linear map $\rho:W\to\Bbb{F}$, there isn’t a unique was to extend it to a linear map $\tilde{\rho}:V\to \Bbb{F}$. This last part seems pretty intuitive to me, so I’m not sure what the issue seems to be.
One way of ‘quantifying’ this non-uniqueness is by the dimension of this space, and one way to calculate this dimension ‘cleanly’ (i.e without too many invocations of bases) is by the universal property for quotient spaces, which tells us that we have an isomorphism \begin{align} W^0\cong \text{Hom}(V/W,\Bbb{F})=(V/W)^*. \end{align} So, these spaces have the same dimension: \begin{align} \dim (W^0)=\dim((V/W)^*)=\dim(V/W)=\dim V-\dim W=n-k, \end{align} which $0$ if and only if $W=V$.
A somewhat similar and related problem is the fact that given a subspace $W\subset V$, there is in general no unique complementary subspace (i.e a subspace $U$ such that $V=W\oplus U$)… also keep in mind that $\dim U=\dim V-\dim W=n-k$.
If now you specialize to say $\Bbb{F}=\Bbb{R}$ and $V$ having a (positive-definite) inner product then yes every subspace $W\subset V$ has a ‘distinguished’ complementary subspace, denoted $W^{\perp}$, and called the orthogonal complement; we now have an orthogonal direct sum decomposition $V=W\oplus W^{\perp}$. But still, keep in mind that even with the orthogonal complement $W^{\perp}$ at hand, there is in general no ‘preferred basis’ (even if you demand the basis be orthonormal) simply because $\dim (W^{\perp})$ is generally strictly larger than $1$. In other words, there are no ‘preferred’ choices of unit normals to $W$. Ok, so let’s say we fix $k=n-1$, i.e a codimension-1 subspace $W$ (i.e a hyperplane in $V$ through the origin). Then, $W^{\perp}$ will be 1-dimensional, but of course, we still do not have a unique choice of unit normal to $W$, because there are exactly two of them. You need something else (such as a suitable choice of orientation) if you want to be able to pick one out of the two as the ‘natural’ one.
In the case of Lorentzian signature (or more generally other pseudo-Euclidean signatures), we of course have the problem that although every subspace $W\subset V$ has a unique ‘orthogonal’ space $W^{\perp}$ of dimension $\dim V-\dim W$, it may happen that $W,W^{\perp}$ are not complementary (this is what happens in SR/GR with null hyperplanes/hypersurfaces). One needs to restrict to special classes of subspaces $W$ in order to get an orthogonal complement (e.g spacelike or timelike subspaces)… but anyway this is a secondary issue I think to what you’re asking.