(1) Which cost function are you thinking off? (2) As there is lots of choice for $T$ the answer seems to be $(*) = \inf_{S:\mathcal{N}\to\mathcal{N} \int c(S(y),y)d\nu(y)$ if the dimensions of the manifolds agree.

CONTINUED: Note that this also means the counterexample is true if $\kappa$ is not a delta measure and $X=\{x_0\}$. Hence the only case left is all $\kappa$s are mapped to the same delta measure $\delta_{y_0}$. In this case $S_\kappa$ is obviously onto as the image is just the trivial and unique coupling of $\delta_{y_0}$ with itself.

The counterexample works when there are at least two measures with different image and where $\kappa$ is a delta measure on a set of full measure of those. Observe that two delta measure have the same image if and only if $\kappa(dy|x_1)=\kappa(dy|x_2)$. If the kernel is not a delta measure then the construction works and the image measures of the two deltas have uncountably many couplings (there are many self-couplings for non-delta measures).

$p_2(\Gamma)$ is the projection to the second factor, i.e. $p_2:(x,y)\mapsto y$. The rough idea of the construction: Look at all initial points of a given final point $y$. If they all lie into a "common" direction then one can move $y$ towards that direction to decrease the transport a bit. Call the moved point $y'$. If everything is transported as before with $y$ replaced by $y'$ then the transport cost is smaller though possibly not optimal (in many cases it is optimal but in a few it is not).

In order to be able to state your question in a Polish space you need to assign a metric. After thinking over night, I believe the second counterexample can be adjusted to any $\mu$. The idea is to exhaust a small ball around $x$ by open sets which become eventually dense but have arbitrary small $\mu$-measure (this is the sequence $x_n$). Picking some ball disjoint from the chosen one (this corrensponds to $x_1$), it's possible to get the same counterexample.

Sorry, it's just a typo. It should have been $d_H$. Answer is now adjusted. The examples show that there can be neither upper nor lower bounds for d_H w.r.t. to the Wasserstein distances (except for $p=\infty$ with the natural upper bound).