$dl'$ is equivalent to "$d|\mathbf{r}|$", it is essentially a "scalar length measure".
The electrodynamics integral you wrote here is a vector-valued integral, so no dotting happens. If you use a linear coordinate system, it may be evaluated as three scalar line integrals, one for each coordinate. Vector valued integrals cannot really be evaluated using a curved coordinate system.
Your integral should be evaluated as follows: $\mathbf{r}'$ is actually a curve $\mathbf{r}'(t)$ (I am not going to use ' as a notation for derivatives), and your integral is $$\mathbf{E}(\mathbf{r})=\kappa\int_\gamma\frac{\lambda(\mathbf{r}'(t))}{|\mathbf{r}-\mathbf{r}'(t)|^2}\frac{\mathbf{r}-\mathbf{r}'(t)}{|\mathbf{r}-\mathbf{r}'(t)|}\left|\frac{d\mathbf{r}'(t)}{dt}\right|dt. $$
Note that whilst I do not know the full context, from the form of the integral I can infer that only $\mathbf{r}'$ is parametrized by $t$. You are asking the value of $\mathbf{E}$ at a single point ($\mathbf{r}$), and this value is given by integrating the $\mathbf{r}'$ variable over a curve.
Edit: To answer your comment, you are almost correct.
Now that I know the full case, you also need to parametrize the $\mathbf{r}$ variable, but only because you are interested in values of $E$ solely on the $z$ axis!
Let us calculate the relevant vector quantities:
$$\mathbf{r}=\mathbf{r}(z)=z\mathbf{e}_z$$ (I am using the generic $z$ as a parameter instead of $b$, since I think this way it is clearer.) $$\mathbf{r}'(t)=a\cos(t)\mathbf{e}_x+a\sin(t)\mathbf{e}_y, $$ $$ \mathbf{r}(z)-\mathbf{r}'(t)=-a\cos(t)\mathbf{e}_x-a\sin(t)\mathbf{e}_y+z\mathbf{e}_z, $$ $$|\mathbf{r}(z)-\mathbf{r}'(t)|^2=a^2+z^2, $$ $$ \frac{d\mathbf{r}'(t)}{dt}=-a\sin(t)\mathbf{e}_x+a\cos(t)\mathbf{e}_y, $$ $$\left|\frac{d\mathbf{r}'(t)}{dt}\right|=a. $$
Now the integral is (since $\lambda$ is constant) $$ \mathbf{E}(\mathbf{r}(z))=\frac{\lambda}{4\pi\epsilon_0}\int_\gamma\frac{1}{a^2+z^2}\left(-\frac{a\cos(t)}{\sqrt{a^2+z^2}}\mathbf{e}_x-\frac{a\sin(t)}{\sqrt{a^2+z^2}}\mathbf{e}_y+\frac{z}{\sqrt{a^2+z^2}}\mathbf{e}_z\right)a\ dt=\\\frac{\lambda}{4\pi\epsilon_0}\frac{a}{(a^2+z^2)^{3/2}}\int_0^{2\pi}z\ dt\mathbf{e}_z=\frac{\lambda}{2\epsilon_0}\frac{az}{(a^2+z^2)^{3/2}}\mathbf{e}_z, $$ where the $\mathbf{e}_x$ and $\mathbf{e}_y$ integrals are annihilated because $\sin(t)$ and $\cos(t)$ are integrated on their full period.
Best Answer
For a point charge, the electric field has a radial component only
$$\mathbf E = \frac{kQ}{r^2}\hat{\mathbf r}$$
Thus, the dot product of the electric field and the infinitesimal displacement vector is
$$\mathbf E \cdot d\mathbf l = \frac{kQ}{r^2}\hat{\mathbf r}\cdot \left( dr\;\hat{\mathbf r} + r\;d\theta \; \hat{\boldsymbol {\theta}} + r\;\sin \theta \;d\phi \;\hat{\boldsymbol{ \phi}}\right) = \frac{kQ}{r^2}\;dr$$
so, only displacement in the radial direction contributes to the line integral.
If the start and end radial coordinates are the same, the line integral is zero.