I thought about this once and came up with an "answer" of sorts.
Notation: Given $z=a+ib$ and $w = c+id$ in $\mathbb{C}$, I will write $z \cdot w$ for the Euclidean dot product $ac + bd$. Ordinary multiplication of complex numbers will be denoted by juxtaposition.
The basic observation is, given $z,w \in \mathbb{C}$, the quantity $\overline z w$ is related to the dot product as follows:
\begin{align*}
\mathrm{Re}(\overline z w) = z \cdot w && \mathrm{Im}(\overline z w) = (iz) \cdot w.
\end{align*}
Note that $iz$, geometrically speaking, is $z$ rotated counterclockwise by 90 degrees.
Now, rather than interpret $\int_\gamma f$, let us interpret $I = \int_\gamma \overline f$. As you noted,
$$ I= \int_0^1 \overline{f(\gamma(t))} \gamma'(t) \ dt,$$
and so
\begin{align*}
\mathrm{Re}(I) = \int_0^1 f(\gamma(t)) \cdot \gamma'(t) \ dt &&
\mathrm{Im}(I) = \int_0^1 (if(\gamma(t))) \cdot \gamma'(t) \ dt
\end{align*}
or, in the notation of line integrals of vector fields,
\begin{align*}
\mathrm{Re}(I) = \int_\gamma f(z) \cdot dz &&
\mathrm{Im}(I) = \int_\gamma (if(z)) \cdot dz.
\end{align*}
So, for instance, thinking of $f$ as a force field,
- the real part of $\int_\gamma \overline f$ is the work done traveling along $\gamma$ through the force field $f$,
- the imaginary part of $\int_\gamma \overline f$ is the work done traveling along $\gamma$ through the force field $if$.
Note $if$ is just the force field $f$ rotated counterclockwise by 90 degrees.
Hope this helps...
Added: I thought I would add a simple example.
Put $f(z) = \overline{1/z}$. As you are no doubt aware, if the contour $\gamma$ is closed and avoids the origin, then $\int_\gamma \overline f = \int_\gamma \frac{1}{z}$ is equal to $2 \pi i$ times the winding number of $\gamma$.
In this example, the vector field $f(z) = \overline{1/z}$ is the gradient of the scalar potential $V(x,y) = \frac{1}{2} \log (x^2+y^2)$, so no work is done going around a contour in this field. The field $f$ looks like this:
On the other hand, the 90 degree rotated field $if$ looks like this:
Added: I just learned the proper name for this interpretation. The complex conjugate $\overline f$, viewed as a vector field, is called the "Polya vector field of $f$". As I mentioned, the real part of $\int_\gamma f$ is the work done by the Polya vector vield on a particle as it travels along the curve $\gamma$. An alternative way to think of the imaginary part of $\int_\gamma f$ is as the flux of the Polya vector field through the oriented curve $\gamma$. These ideas are explained in Chapter 11 of Tristan Needham's book "Visual Complex Analysis".
Best Answer
So, the thing about those path integrals - the reason you chose that path - is that some will tend to variations on the integral we want to evaluate, and the others will go to zero. That is how we use the residue theorem to evaluate integrals; we don't start with a closed path, so we come back around to close it with pieces we can understand.
Adding up the pieces, we get a limit that's a sum of two constant multiples of the integral we're trying to evaluate. In order to solve for that integral, we need a second evaluation of the contour integral - and that calls for the residue theorem. There's exactly one pole in the rectangle enclosed, at $\pi i$.
Can you take it from here?