Given a position function of a particle: $$ \mathbf r=r\,\hat{\mathbf r}\left(\theta\right), $$ where $\hat{\mathbf r}(θ)$ is the polar unit vector, to find the velocity, we take the derivative which results in:
$$ \mathbf v = \frac{\mathrm d\mathbf r}{\mathrm dt} = \frac{\mathrm dr}{\mathrm dt} \hat{\mathbf r}(θ) + r\frac{\mathrm d\hat{\mathbf r}(θ)}{\mathrm dt}.$$
What does each term in the velocity equation mean?
You've made things a bit more complicated than they need to be by introducing the unspecified parameter $\theta$. So let's get rid of it and write $$\boldsymbol r(t) = r(t)\,\hat{\boldsymbol r}(t)$$ Taking the derivative with respect to time yields $$\frac{d\,\boldsymbol r(t)}{dt} = \frac{dr}{dt}\,\hat{\boldsymbol r}(t) + r(t)\frac{d\,\hat{\boldsymbol r}(t)}{dt}$$ The first term is the radial component of velocity and the second is the tangential component.
Given any two vectors, one can always express the second vector as comprising two components, one that is parallel to the first and another that is orthogonal to the first. This is exactly what is happening here: The first term ($\dot r\,\hat {\boldsymbol r}$) is necessarily parallel to the position vector while the second ($r \dot{\hat {\boldsymbol r}}$) is necessarily orthogonal to the position vector.
Now I'll reintroduce the intermediate parameter $\theta$. Suppose that $\boldsymbol r$ depends directly on $\theta$ only: $$\boldsymbol r(\theta) = r(\theta)\,\hat{\boldsymbol r}(\theta)$$ For example, $$\begin{aligned} r(\theta) &= \frac{a(1-e^2)}{1-e\cos\theta} \\ \hat{\boldsymbol r}(\theta) &= \cos\theta\,\hat{\boldsymbol x}+\sin\theta\,\hat{\boldsymbol y} \end{aligned}$$ Assuming $\theta$ varies with time, this describes an elliptical orbit about the origin. Differentiating with respect to time yields $$\frac{d\,\boldsymbol r(t)}{dt} = \frac{dr}{d\theta}\frac{d\theta}{dt}\,\hat{\boldsymbol r}(\theta) + r(\theta)\frac{d\,\hat{\boldsymbol r}(\theta)}{d\theta}\frac{d\theta}{dt}$$ As was the case where $r$ depended directly on $t$ only, the first term is necessarily parallel to the position vector while the second is necessarily orthogonal to the position vector.
