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A crude dimension count now shows that, when $\dim V = d > 3$, the union of all the $W(V_1,V_2)$ in $W(V)$ when $V_1$ and $V_2$ are orthogonal complements in $V$ is a proper subvariety of $W(V)$ and that, moreover, the generic element in $W(V_1,V_2)$ does not lie in $W(V'_1, V'_2)$ for any other distinct splitting of $V$.

A crude dimension count now shows that, when $\dim V = d > 3$, the union of all the $W(V_1,V_2)$ in $W(V)$ when $V_1$ and $V_2$ are orthogonal complements in $V$ is a proper subvariety of $W(V)$ and that, moreover, usually, the generic element in $W(V_1,V_2)$ does not lie in $W(V'_1, V'_2)$ for any other distinct splitting of $V$. (The reason for the 'usually' is that there is an exception when $2=\dim V_1\le \dim V_2$ and $1 = \dim V'_1<\dim V'_2$).

In dimensions above 3: While an exhaustive analysis is probably rather complicated, some general remarks about the dimensions above $3$ will give a sense of what is involved:

First, it is reasonable to ask what one can say about the Weyl curvature of a product metric. First, a few definitions: If $V$ is a vector space endowed with a positive definite inner product, let $R(V)\subset S^2\bigl(\Lambda^2(V^*)\bigr)$ denote the space of Riemann curvature tensors, and let $R(V) = W(V) \oplus Rc(V)$ be the $O(V)$-invariant direct sum decomposition of $R(V)$ into the space $W(V)$ of Weyl curvatures and the $O(V)$-invariant complement $Rc(V)$. If $d\ge 3$ is the dimension of $V$, then $$ \dim R(V) = \frac{d^2(d^2{-}1)}{12} \qquad\text{and}\qquad \dim W(V) = \frac{d(d{+}1)(d{+}2)(d{-}3)}{12}\ . $$ Now, if $V_1$ and $V_2$ are two vector spaces with dimensions $d_1>0$ and $d_2>0$, respectively, with positive definite inner products, and $V_1\oplus V_2$ is given the natural inner product structure, then one has a natural mapping $$ R(V_1)\oplus R(V_2)\longrightarrow R(V_1\oplus V_2)\longrightarrow W(V_1\oplus V_2), $$ and, except when $d_1=d_2=1$, this mapping has a kernel of dimension $1$. (This corresponds to the fact that a nontrivial Riemannian product of dimension $3$ or more is conformally flat in only two cases: $\mathbb{R}\times M_c$, where $M$ has constant sectional curvature $c$, and $M_{-c}\times N_c$, where $M_{-c}$ has constant sectional curvature $-c$ and $N_c$ has constant sectional curvature $c$.) Let the image of the above mapping be denoted $W(V_1,V_2)\subset W(V_1\oplus V_2)$

A crude dimension count now shows that, when $\dim V = d > 3$, the union of all the $W(V_1,V_2)$ in $W(V)$ when $V_1$ and $V_2$ are orthogonal complements in $V$ is a proper subvariety of $W(V)$ and that, moreover, the generic element in $W(V_1,V_2)$ does not lie in $W(V'_1, V'_2)$ for any other distinct splitting of $V$.

It follows that, most of the time, in dimensions $d>3$, you'll be able to rule out a Riemannian metric $g$ on a $d$-manifold $M$ being conformal to a product, just by examining the Weyl curvature. Moreover, most of the time that the Weyl curvature of $g$ does actually happen to, pointwise, satisfy the condition of being the Weyl curvature of a product, there will be a unique splitting of the tangent bundle $TM = D_1\oplus D_2$ such that the Weyl curvature of $g$ takes values in the bundle $W(D_1,D_2)$.

For example, when $d=4$, the set of Weyl curvatures in $W(V)$ (a vector space of dimension $10$) that are Weyl curvatures of a product of surfaces is a cone of dimension $5$ that is smooth away from the origin. If you have a Riemannian $4$-manifold whose Weyl curvature is nonvanishing but does take values in this cone, then there is only one splitting of the tangent bundle into $2$-plane bundles that could possibly be the tangents to the leaves of a product structure that could support a product metric conformal to your given metric.

Once you have the only possible splitting $TM = D_1\oplus D_2$ that could work, going the rest of the way is easy: First, the two plane fields $D_i$ must be integrable. Second, when you write $g = g_1 + g_2$ where $g_i$ is positive definite on $D_i$ and has $D_{3-i}$ as null space, then the Lie derivative of $g_i$ with respect to any vector field tangent to $D_{3-i}$ has to be a multiple of $g_i$, or else it cannot work. Third, letting $d_i$ denote the rank of $D_i$, if you now let $\Omega_i$ be a decomposable $d_i$-form that has $D_{3-i}$ as its kernel and is the $g_i$-induced volume form on $D_i$ (this specifies $\Omega_i$ up to a sign), then there will exist a unique $1$-form $\lambda$ such that $$ d\Omega_i = d_i\ \lambda\wedge \Omega_i $$ (this follows from the integrability that was the first condition); the final local requirement is that this $\lambda$ must be closed, i.e., $d\lambda=0$. If you have this, then, setting $\lambda = d f$ for some function $f$, one sees that $$ e^{-2f}g = e^{-2f}g_1 + e^{-2f}g_2 $$ and each of the scaled metrics $e^{-2f}g_i$ is well-defined on the $d_i$-dimensional leaf space of the integrable plane field $D_{3-i}$. Thus, $g$ is conformal to a product metric.

Of course, there will be metrics whose Weyl curvatures can be written as split in more than one way (for example, the conformally flat metrics, which have zero Weyl curvature), and a more subtle analysis would need to be done for those, but these will be relatively rare and the analysis is liable to be somewhat messy and dimension dependent. I wouldn't undertake to tackle these special cases without a good reason.

Meanwhile, in dimension $3$, this rank restriction on $C\bigl([g]\bigr)$ is not sufficient to make $[g]$ conformal to a product. To see the further restrictions, suppose that $C\bigl([g]\bigr)$ has rank $2$ everywhere on the $3$-manifold $P$. Then it follows by algebra that there will be a coframing $\eta = (\eta_1,\eta_2,\eta_3)$, unique up to some changes of sign and switching $\eta_1$ and $\eta_2$, so that $[g] = [{\eta_1}^2+{\eta_2}^2+{\eta_3}^2]$ and so that $$ C\bigl([g]\bigr) = \bigl(2\ \eta_1{\circ}\eta_2\bigr)\otimes (\eta_1\wedge\eta_2\wedge\eta_3)^{1/3} $$ Comparing this with the above formula in the case of a known product, in which the two 'factors' of $C\bigl([g]\bigr)$, namely $dt$ and $*dK$, are integrable), one sees that a further necessary condition is that $\eta_1$ and $\eta_2$ be integrable, i.e., $\eta_i\wedge d\eta_i = 0$ for $i = 1, 2$. Because of the Bianchi identities, this turns out to be only one further condition (this one of $4$th order) on $[g]$.

Indeed, the Bianchi identities for the conformal structure in this coframing can be expressed as follows: There exist functions $p_1,p_2,p_3, q, r$ such that the following formulae hold, with $\pi = p_1\eta_1+p_2\eta_2+p_3\eta_3$ $$ \begin{aligned} d\eta_1 &= \pi\wedge\eta_1 + q\,\eta_3\wedge\eta_1 + 3p_2\,\eta_1\wedge\eta_2\\ d\eta_2 &= \pi\wedge\eta_2 + q\,\eta_2\wedge\eta_3 - 3p_1\,\eta_1\wedge\eta_2\\ d\eta_3 &= \pi\wedge\eta_3 + 3p_2\,\eta_2\wedge\eta_3 - 3p_1\,\eta_3\wedge\eta_1 + r\,\eta_1\wedge\eta_2\\ \end{aligned} $$

Now, if the first local product structure is to work, then the quadratic form $Q_1 = {\eta_2}^2+{\eta_3}^2$ will have to be invariant up to multiples under the flow of the vector field $X_1$ dual to $\eta_1$. One sees immediately that this is equivalent to the conditions $r = p_1 = 0$. Similarly, if the first local product structure is to work, then the quadratic form $Q_2 = {\eta_1}^2+{\eta_3}^2$ will have to be invariant up to multiples under the flow of the vector field $X_2$ dual to $\eta_2$, which is equivalent to the conditions $r = p_2 = 0$. Either way, this represents two more $4$th order equations on $[g]$.

Finally, suppose that $r=p_1=0$. Then the necessary and sufficient local condition is that $d(p_2\ \eta_2 + (p_3{+}q)\ \eta_3)=0$ (which represents two $5$th order equations on $[g]$), for, then setting $d f = p_2\ \eta_2 + (p_3{+}q)\ \eta_3$, one sees that $\eta_1 = e^f\ dx_1$ for some function $x_1$ and that $f$ is constant on the integral curves of $X_1$, so that $$ e^{-2f} g = {dx_1}^2 + e^{-2f} Q_1 $$ where $e^{-2f} Q_1$ is a well-defined metric on surface that is the space of integral curves of $X_1$. There is a completely analogous case when $r=p_2=0$, of course.

Just one more remark: If you have $r = p_1 = p_2 = 0$ and pursue the analysis, you can show that the conformal structures that are conformal to a product in two distinct nontrivial ways (and are not conformally flat) can be written in the form $$ [g] = \bigl[f_1(x_3)\ {dx_1}^2 + f_2(x_3)\ {dx_2}^2 + {dx_3}^2\bigr] $$ where $f_i$ for $i=1,2$ are positive functions of one variable. Thus, the nonflat conformal structures that have more than one 'conformal product structure', even locally, depend on two arbitrary functions of one variable.

Meanwhile, in dimension $3$, this rank restriction on $C\bigl([g]\bigr)$ is not sufficient to make $[g]$ conformal to a product. To see the further restrictions, suppose that $C\bigl([g]\bigr)$ has rank $2$ everywhere on the $3$-manifold $P$. Then it follows by algebra that there will be a coframing $\eta = (\eta_1,\eta_2,\eta_3)$, unique up to some changes of sign and switching $\eta_1$ and $\eta_2$, so that $[g] = [{\eta_1}^2+{\eta_2}^2+{\eta_3}^2]$ and so that $$ C\bigl([g]\bigr) = \bigl(2\ \eta_1{\circ}\eta_2\bigr)\otimes (\eta_1\wedge\eta_2\wedge\eta_3)^{1/3} $$ Comparing this with the above formula in the case of a known product, in which the two 'factors' of $C\bigl([g]\bigr)$, namely $dt$ and $*dK$, are integrable, one sees that a further necessary condition is that $\eta_1$ and $\eta_2$ be integrable, i.e., $\eta_i\wedge d\eta_i = 0$ for $i = 1, 2$. Because of the Bianchi identities, this turns out to be only one further condition (this one of $4$th order) on $[g]$.

Indeed, the Bianchi identities for the conformal structure in this coframing, plus the above integrability conditions, can be expressed as follows: There exist functions $p_1,p_2,p_3, q_1, q_2$ such that the following formulae hold $$ \begin{aligned} d\eta_1 &= q_1\,\eta_3\wedge\eta_1 + p_1\,\eta_1\wedge\eta_2\\ d\eta_2 &= q_2\,\eta_2\wedge\eta_3 + p_2\,\eta_1\wedge\eta_2\\ d\eta_3 &= 2p_1\,\eta_2\wedge\eta_3 + 2p_2\,\eta_3\wedge\eta_1 + 2p_3\,\eta_1\wedge\eta_2\\ \end{aligned} $$

Now, if the first local product structure is to work, then the quadratic form $Q_1 = {\eta_2}^2+{\eta_3}^2$ will have to be invariant up to multiples under the flow of the vector field $X_1$ dual to $\eta_1$. Since the above formulae imply $$ L_{X_1}({\eta_2}^2+{\eta_3}^2) = 2p_2\ ({\eta_2}^2-2{\eta_3}^2) + 4p_3\ \eta_2{\circ}\eta_3\ , $$ one sees that this condition is $p_2 = p_3 = 0$. Note that these conditions imply that $\eta_2$ and $\eta_3$ separately are invariant under the flow of $X_1$. Similarly, if the first local product structure is to work, then the quadratic form $Q_2 = {\eta_1}^2+{\eta_3}^2$ will have to be invariant up to multiples under the flow of the vector field $X_2$ dual to $\eta_2$, which is $p_1 = p_3 = 0$. Either way, this represents two more $4$th order equations on $[g]$.

Finally, suppose that $p_2=p_3=0$. Then the necessary and sufficient local condition is that $d(q_1\ \eta_3 - p_1\ \eta_2)=0$ (which represents two $5$th order equations on $[g]$), for, then setting $d f = q_1\ \eta_3 - p_1\ \eta_2$, one sees that $\eta_1 = e^f\ dx_1$ for some function $x_1$ and that $f$ is constant on the integral curves of $X_1$, so that $$ e^{-2f} \bigl({\eta_1}^2+{\eta_2}^2+{\eta_3}^2\bigr) = {dx_1}^2 + e^{-2f} \bigl({\eta_2}^2+{\eta_3}^2\bigr) $$ where $e^{-2f}\bigl({\eta_2}^2+{\eta_3}^2\bigr)$ is a well-defined metric on the surface that is the space of integral curves of $X_1$. Now, the right hand side of the above equation is visibly a product metric. There is a completely analogous case when $p_1=p_3=0$, for the other possible local product structure.

Just one more remark: When $p_1 = p_2 = p_3 = 0$, one can pursue this analysis and show that the conformal structures that are conformal to a product in two distinct nontrivial ways (but are not conformally flat) can be written in the form $$ [g] = \bigl[f_1(x_3)\ {dx_1}^2 + f_2(x_3)\ {dx_2}^2 + {dx_3}^2\bigr] $$ where $f_i$ for $i=1,2$ are positive functions of one variable. Thus, the nonflat conformal structures that have more than one 'conformal product structure', even locally, depend only on two arbitrary functions of one variable.