First, let me point out that the OP's suggested generalization to arbitrary target manifolds $Y$ of asking that $f^*\omega = g^*\omega$ for *all* $k$-forms on $Y$, is *not* equivalent to the question about equality of $k$-by-$k$ minors.

To see this, consider the simplest case, $X = Y=\mathbb{R}^1$ and $k=1$. The general $1$-form is of the form $\omega = h(x)\,\mathrm{d}x$ (where $x:\mathbb{R}\to\mathbb{R}$ is the identity function), and $f^*\omega = g^*\omega$ for all $1$-forms $\omega$ is equivalent to
$$
h\bigl(f(x)\bigr)f'(x) = h\bigl(g(x)\bigr)g'(x)
$$
for all functions $h:\mathbb{R}\to\mathbb{R}$, which clearly implies that $f(x) = g(x)$ for all $x\in\mathbb{R}$. Meanwhile, $f^*\mathrm{d}x = g^*\mathrm{d}x$ only implies $f'(x) = g'(x)$, which *is* the same as requiring that all the $1$-by-$1$ Jacobian minors of $f$ and $g$ are equal, which does imply that they differ by an additive constant.

Instead, I think that the natural generalization is that one endows the $n$-manifold $Y$ with a *coframing*, i.e., a basis $\omega = (\omega^1,\ldots,\omega^n)$ of the $1$-forms on $Y$ (which, of course, requires that $Y$ be parallelizable, in fact, $\omega:TY\to\mathbb{R}^n$ defines a linear isomorphism $\omega_y:T_yY\to\mathbb{R}^n$ for each $y\in Y$) and then require that
$$
f^*(\omega^{i_1}\wedge\cdots\wedge\omega^{i_k})
= g^*(\omega^{i_1}\wedge\cdots\wedge\omega^{i_k})
$$
for all $i_1<i_2<\cdots <i_k$.

Even this will generally force some 'finite' equations on $f$ and $g$ if $\omega$ is chosen generally (and the dimension of $X$ is greater than $k$). For a multi-index $I = (i_1,\ldots,i_p)$ with $1\le i_1<i_2<\cdots<i_p\le n$, write $|I|=p$ and $\omega^I$ for $\omega^{i_1}\wedge\cdots\wedge\omega^{i_p}$. Then there will be functions $h^I_J$ on $Y$ such that
$$
\mathrm{d}\omega^I = \sum_{|J|=|I|{+}1} h^I_J\,\omega^J.
$$
If the functions $h^I_J$ are not constant, and $f$ and $g$ both have differential rank at least $k{+}1$, then applying the exterior derivative to the equations $f^*\omega^I = g^*\omega^I$ for $|I|=k$ will generally force some relations on the functions $f^*h^I_J$ and $g^*h^I_J$. One can avoid this 'problem' by assuming that the functions $h^I_J$ be constants. For example, when one takes the standard coordinate coframing on $Y = \mathbb{R}^n$, one has $h^I_J=0$. More generally, if $Y$ is a Lie group and the $\omega^i$ are a basis for the left-invariant forms on $Y$, then the $h^I_J$ are constants. In this latter case, when $X$ is connected, one will have $f^*\omega^i=g^*\omega^i$ for all $i$ if and only if $g = \lambda_y\circ f$ where $\lambda_y:Y\to Y$ is left multiplication by $y\in Y$ (regarded as a Lie group). So I think that this is the natural generalization of the OP's case of $\mathbb{R}^n$.

Second, let me point out that, if the differential ranks of $f$ and $g$ are both less than $k$ at every point, then, of course, $f^*\omega = g^*\omega=0$ for all all $k$-forms $\omega$ on $Y$, so there is no further condition implied than the rank condition. Thus, to get an interesting problem, one must assume that the differential ranks of $f$ and $g$ are *at least* $k$ in order to get an interesting theory.

Once one assumes this, there are some reasonable things to say. For example, if one assumes that the differential ranks of $f$ and $g$ are both at least $k$, then the condition $f^*\omega^I = g^*\omega^I$ for all $|I|=k$ implies that $\mathrm{ker}(f'(x)) = \mathrm{ker}(g'(x))\subset T_xX$ and $\omega_x\bigl(f'(x)(T_x)\bigr) = \omega_x\bigl(g'(x)(T_x)\bigr) \subset\mathbb{R}^n$ for all $x\in X$. Moreover, if one sets $K_x = \mathrm{ker}(f'(x)) = \mathrm{ker}(g'(x))\subset T_xX$ and $Q_x = \omega_x\bigl(f'(x)(T_x)\bigr) = \omega_x\bigl(g'(x)(T_x)\bigr) \subset\mathbb{R}^n$, then the induced isomorphisms
$$
[f'(x)],[g'(x)]:T_xX/K_x\to Q_x
$$
satisfy, for all $x\in X$,
$$
E_k\bigl([f'(x)]\bigr) = E_k\bigl([g'(x)]\bigr):E_k(T_xX/K_x)\to E_k(Q_x),
$$
where $E_k$ is the '$k$th-exterior power' functor on the category of vector spaces and linear maps.

When the differential ranks of $f$ and $g$ are equal to $k$, this is not a very strong condition, so $f$ and $g$ need not be closely related. For example when $m=n=k$, and $X = Y = \mathbb{R}^k$, then the OP's condition on $f$ and $g$ reduces to the assumption that the $f$- and $g$- pullbacks of the standard volume form on $Y$ are equal, and, of course, there are many such pairs $f$ and $g$ besides the translations.

When the differential ranks of $f$ and $g$ are both greater than $k$, though, this is a much stronger condition. In fact, one finds that, when $k$ is odd, this is equivalent to $[f'(x)] = [g'(x)]$ while, when $k$ is even, this is equivalent to $[f'(x)] = \pm[g'(x)]$. Once one is in this situation, at least when one assumes that the differential ranks of $f$ and $g$ are constant, the Cartan equivalence method can be applied. For example, one has the following result:

**Proposition:** Suppose that $f,g:\mathbb{R}^m\to\mathbb{R}^n$ are smooth maps of constant differential rank *greater* than $k\ge1$ and suppose that all their corresponding $k$-by-$k$ Jacobian minors are equal. If $k$ is odd, then $g = c + f$ where $c\in\mathbb{R}^n$ is a constant. If $k$ is even, then $g = c \pm f $ where $c\in\mathbb{R}^n$ is a constant.