I have problem formulation saying
$ q:X\times W \rightarrow \mathbb{R} $
is convex wrt. its second argument $\forall\,x\in X$.
How is different from saying q is convex?
And does this mean $\frac{\partial^2q(x,w)}{\partial w^2}\geq 0 \, \forall\,x\in X$?
On
From a conceptual point of view, $X\times W$ must not be a convex set, so that it is nonsense to talk about the convexity of the function $q$. Indeed, for all $(x_1,w_1),(x_2,w_2)\in X\times W$ and $t\in[0,1]$, one must have: $$(1-t)(x_1,w_1)+t(x_2,w_2)\in X\times W.$$ Your assumption ensures that for all $x\in X$, $q(x,\cdot)$ is convex and hence if $q(x,\cdot)$ is twice differentiable: $$\frac{\mathrm{d}q(x,\cdot)}{\mathrm{d}w}(w)=\frac{\partial q}{\partial w}(x,w)\geqslant 0.$$
What it means when it says "$q$ is convex w.r.t. $w$" is that for any constant value of $x=x_{0}$ (i.e. only considering $q$ as a function of $w$), $q$ is convex. This is not the same as just saying $q$ is convex, since in the former case we have qualified our statement*.
Consider the function $q(x,w) = w^{2}-x^{2}$. If we hold $x$ equal to some constant $x_{0}$, then $q(x_{0},w) = w^{2}-x_{0}^{2} = w^{2}-\mathrm{constant}$, and so $q$ is convex w.r.t $w$. But similarly, if we hold $w$ constant, we see that we have $q(x,w_{0}) = \mathrm{constant}-x^{2}$, so $q$ is in fact concave w.r.t. $x$ and hence not overall convex.
To test for overall convexity, you would need to prove:
$$tq(x_{1},w_{1})+(1-t)q(x_{2},w_{2}) \ge q(tx_{1}+(1-t)x_{2},tw_{1}+(1-t)w_{2}) \quad \forall x_{1},x_{2} \in X, \forall w_{1},w_{2} \in W, \forall t \in [0,1]$$
For convexity w.r.t. a single parameter, you are correct in considering the second derivative (provided the function is twice-differentiable).
* Similarly, one could say that $\sin{x}$ is convex on the interval $[\pi,2\pi]$, which it is, though it is not convex overall - the difference again being that the first statement is qualified.