Let's dig into the inexplicit steps of this calculation!
\begin{align} \mathsf E(X+Y)&=E[Z] && \text{by definition of Z = X+Y}\\\\ &=\int_\Bbb R z~p(z) ~ \mathsf d z && \text{by definition of the expected value}\\ &=\int_\Bbb R z ~ \Big( \int_\Bbb R p(z \cap y) ~ dy \Big) \mathsf d z && \text{by the law of total probability}\\ &=\int_\Bbb R z ~ \Big( \int_\Bbb R p(z | y) ~ p(y) ~ dy \Big) \mathsf d z && \text{by definition of the conditional probability}\\ &=\int_\Bbb R z ~ \Big( \int_\Bbb R p(z - y | y) ~ p(y) ~ dy \Big) \mathsf d z && \text{since p(Z=x+y|Y=y) = p(X=x|Y=y)}\\ &=\int_\Bbb R \Big( \int_\Bbb R z ~ p(z - y | y) ~ p(y) ~ dy \Big) \mathsf d z && \text{by linearity of the integral}\\ &=\int_\Bbb R \Big( \int_\Bbb R (x + y) ~ p(x | y) ~ p(y) ~ dy \Big) \mathsf d x && \text{by substituting $x = z - y$ (note that $dx = dz$)} \end{align}
By linearity of the integral, this last line is equal to $$\int_\Bbb R x \Big( \int_\Bbb R p(x | y) p(y) dy \Big) \mathsf d x + \int_\Bbb R \Big( \int_\Bbb R y ~ p(x | y) p(y) dy \Big) \mathsf d x$$
Let's deal with the first term: \begin{align} \int_\Bbb R x \Big( \int_\Bbb R p(x | y) p(y) dy \Big) \mathsf d x &=\int_\Bbb R x \Big( \int_\Bbb R p(x \cap y) dy \Big) \mathsf d x && \text{conditional prob}\\ &=\int_\Bbb R x p(x) \mathsf d x && \text{total prob}\\ &=E[X] \end{align}
Let's deal with the second term. We'll make the reasonable assumption that X and Y are measurable so as to use Fubini's theorem. This theorem allows us to change the order of integrals: \begin{align} \int_\Bbb R \Big( \int_\Bbb R y ~ p(x | y) ~ p(y) ~ dy \Big) \mathsf d x &=\int_\Bbb R y \Big( \int_\Bbb R p(y) ~ p(x | y) ~ dx \Big) \mathsf d y &&\text{with Fubini's theorem}\\ &=\int_\Bbb R y \Big( \int_\Bbb R p(x \cap y) ~ dx \Big) \mathsf d y &&\text{conditional prob}\\ &=\int_\Bbb R y ~ p(y) ~ \mathsf d y &&\text{total prob}\\ &=E[Y] \end{align}
Because in my point of view I may write \begin{align}\mathsf E(X+Y)&=\int_\Bbb R (x+y)~f_{X+Y}(x+y)\mathsf d(x+y)\\&= \int_\Bbb R (x+y)~f_{X+Y}(x+y)\mathsf (dx+dy) \\&= \int_\Bbb R ~f_{X+Y}(x+y) (xdx+ydx+xdy+ydy)\\&\vdots \end{align}
Since you are integrating once over a given support or space (in this case $\mathbb{R}$), $d(x+y)$ is a confusing notation which gives you the impression that you could split it in two ($dx + dy$), when in fact it is not the case. You can think of $dz$ as the part of these notations that indicates what is the name of the variable that "scans" your support.