Check out Kock's Synthetic Geometry of Manifolds for a nice geometric description of connection forms, curvature forms, and torsion in terms of parallel transport. My mathoverflow answer here gives an explanation of torsion from this point of view.

I'll give a run down for how to intuitively think of the connection form. Let $\nabla$ be a choice of connection (i.e. a parallel transport map) in a bundle $E \to M$. If $x$ and $y$ are infinitesimally close points (to first order, technically) in $M$ and $u$ is in the fibre $E_x$, then we can transport $u$ along the (horizontal lift of) the infinitesimal line segment linking $x$ to $y$ to get an element $\nabla(x,y)u$ in the fibre $E_y$.

In other words, for any pair of infinitesimal neighbours $x, y \in M$ we have a map $$\nabla(x,y): E_x \to E_y.$$
We just require that $\nabla(x,x) = \text{id}_{E_x}$.

If the bundle $E$ has some extra structure, we often want the parallel transport to preserve that structure. For example, for a vector bundle we probably want the maps $\nabla(x,y)$ to be linear, and for a $G$-principal bundle we probably want the maps to be $G$-equivariant. This gives us the notion of a linear connection, and a principal connection, respectively.

Now, lets see where connection forms come in. Say we have a connection $\nabla$ on a (right) $G$-principal bundle $p: P \to M$. We don't need it to be $G$-equivariant, but that would obviously be nice. Anyway, if $u$ and $v$ are infinitesimal neighbours in $P$, then we can project down to $M$ to get a pair of infinitesimal neighbours $p(u)$ and $p(v)$. Let's transport $u$ to the fibre containing $v$ along the corresponding infinitesimal curve to get the element $\nabla(p(u), p(v))u$ which lies in the same fibre as $v$. But the action of $G$ on $P$ is simply transitive, so there is a unique element of $\omega(u,v) \in G$ such that
$$v.\omega(u,v) = \nabla(p(u), p(v))u.$$
The $\omega$ here is just the connection form!

But wait, you might ask, why does $\omega$ take a pair of infinitesimally close points in $P$ as arguments rather than a tangent vector to $P$, and why does it have values in $G$ rather than the lie algebra $\frak{g}$? Well, the answer to the first question should be obvious: if two points in $P$ are infinitesimally close (to first order), they lie on a tangent vector to $P$. The answer to the second question is that $\omega(u,v)$ is always an infinitesimal neighbour of the group identity $e \in G$, so it lies on a tangent vector to $G$ at $e$, and hence an element of the lie algebra $\mathfrak{g}$.

If you draw out a picture of this situation you can see that the connection form is the "vertical part" of the parallel transport on a principal bundle.

It is easy to check that $G$-equivariance of a principal connection (transport map) is equivalent to the usual equivariance of the connection form (with respect to the principal and adjoint action) in this picture.

Now, to interpret the curvature, lets just think about any bundle $\pi: E \to M$ with connection $\nabla$. Let's use the notation $x \sim y$ to indicate that two points are infinitesimal neighbours. If we have three points $x,y,z \in M$ such that $x \sim y$, $y\sim z$ and $z\sim x$. These three define an infinitesimal 2-simplex in $M$. Lets consider the transport around (the boundary of) this simplex:
$$R(x,y,z) = \nabla(z,x) \circ \nabla(y,z) \circ \nabla(x,y): E_x \to E_x$$
If we transport a point $w \in E_x$ around the simplex, we have no guarantee that we end up back where we started. This is precisely the notion of curvature. The curvature measures the extent to which parallel transport around infinitesimal 2-simplices deviates from the identity.

If $E$ is a vector bundle with a linear connection, then $R(x,y,z): E_x \to E_x$ is linear, as expected. If $E$ is a principal bundle, then we can use the same trick we used for the connection form to define the $\frak{g}$-valued curvature form $\Omega(x,y,z)$ from the the curvature $R$.