Geometric meaning of Berezin integration Berezin integration in a Grassmann algebra is defined such that its algebraic properties are analogous to definite integration of ordinary functions: linearity (taking anticommutativity into account), scale invariance and independence from the integration variable. Up to normalization this means that
$$\int f(\theta) d\theta = \int (f_0 + f_1\theta) d\theta = f_1$$
I already have a good geometric picture of Grassmann numbers themselves as elements of an exterior algebra, but I've never understood the geometric meaning of Berezin integration. What exactly are we "summing" and over what domain? Why the strange scaling property $d(a\theta) = a^{-1} d\theta$?
 A: I am quite sure that Berezin integral cannot be interpreted as a sum in any sense. But one can find a geometric picture of Berezin integral.
How should one think about supermanifold geometrically? One way to think about manifold is following. It is a topological space and sheaf of functions (i.e. for each open subset one assign a ring). You may want to consider smooth, complex, algebraic etc manifold. Then those functions should be $C^{\infty}$, holomorphic or algebraic respectively.
(By the way, there are some conditions... you cannot take topological space to be a point and assign a ring of polynomials.)
For supermanifold rings (assigned to each open subspace) are supercommutative (i.e. $\mathbb{Z}/2\mathbb{Z}$ graded with $ab=(-1)^{|a| |b|} ba$). So the point is that if one factor over elements of degree 1, then it is an ordinary manifold. This manifold is called underlying manifold.
This picture recalls a standard picture from algebraic geometry about non-reduced scheme. Let me briefly remind you using an example. Consider intersection of circle $x^2 + y^2 = 1$ and a line $y=1$. This intersection is a point. On the other hand $\mathbb{k}[x,y]/(y-1, x^2+y^2-1) = \mathbb{k}[x]/(x^2)$. So the ring of functions is not $\mathbb{k}$ (as it should be for point). This is a double point. Also one can think about this double point as a fusion of two regular points.
To sum up, in ordinary algebraic geometry there may emerge nilpotents in a ring of functions. One should think about them as infinitesimal thickening of our initial variety.
The same happens with supermanifold. Note that all these odd coordinates are nilpotent. However, I would not recommend you to think about this odd direction as thickening. For example, correct notion of dimension of supermanifold is a number of even coordinates minus number of odd coordinates. I will try to justify this mysterious concept later.
When I try to explain this concept to a physicist, I say that odd coordinates do not correspond to any physical direction. That is why Berezin integral is not a sum.
Construction of supermanifold Let $M$ be a manifold an $E$ be a vector bundle over $M$. Let us define a supermanifold $\Pi E$. As topological space, it is homeomorphic to $M$. But the ring of functions is sections of $\Lambda^* (E^*)$. So locally there are $\dim M$ even coordinates and $\text{rank} E$ odd coordinates. 
Theorem Any supermanifold is (not canonically) isomorphic to $\Pi E$ for some $E$.
Canonical measure on $\Pi T M$. The functions on this space is are differential forms $\theta_i = dx_i$. One can integrate a differential form of highest rank over $M$. We will see that this correspond to canonical measure.
One should think about $d \theta_i$ as a vector field since $\int \theta_j d \theta_i = \delta_{ij}$ (so $d \theta_i$ is dual basis to $\theta_i$). So $d \theta_1 \dots \theta_n dx_1 \dots dx_n$ is canonically defined. 
Let $F(x, \theta)= f_0(x) + \dots + f_n(x) \theta_1 \dots \theta_n$
Then
$$\int_{\Pi TM}  F(x, \theta) d \theta_1 \dots d\theta_n dx_1 \dots dx_n = \int_{M} f_n(x)  dx_1 \dots dx_n$$
So this is standart integration of form of highest rank! 
This picture justifies dimension formula since canonical measure can appear only on a zero-dimensional manifold (do not be too strict to this argument).
Berezin measure. Since any supermanifold is isomorphic to $\Pi E$, let us think about Berezin measure on $\Pi E$. It is section of $\Lambda E^* \otimes \Omega^{\dim M} M$. Integration is following procedure. First of all one use pairing $\Lambda^r E^* \otimes \Lambda^r E \rightarrow \mathbb{k}$. So Berezin measure defines a map
$$\Gamma ( \Lambda^r E ) \rightarrow \Omega^n M $$
Then one just integrate differential form.
This procedure explains geometric meaning of Berezin integral in terms of classical (not super) geometry.
A: *

*User AccidentalFourierTransform has already given a good answer. He is exactly right that Berezin integration $$\int\! d\theta= \frac{\partial }{\partial \theta}\tag{A} $$ is the same as differentiation, which explains the strange scaling property that OP noticed.


*I explain in the first part of my Phys.SE answer here why definition (A) is unique (up to an overall multiplicative normalization factor).


*One thing that one may wonder about is whether it makes sense to define a definite integral$^1$ $$I(\theta_1,\theta_2):=\int_{\theta_1}^{\theta_2}\! d\theta\tag{B}$$ over a Grassmann-interval $[\theta_1,\theta_2]$? I investigate this in the second part of my Phys.SE answer here. While I stop short of claiming a no-go theorem, the various possibilities are rather un-appealing/not useful.


*In particular, it is not useful to think of Berezin integration (A) over the odd line $\mathbb{R}^{0|1}$ as composed of some sort of sum over Grassmann-intervals, cf. OP's question. It's a purely algebraic construction, cf. eq. (A).
References:

*

*Pierre Deligne and John W. Morgan, Notes on Supersymmetry (following Joseph Bernstein). In Quantum Fields and Strings: A Course for Mathematicians, Vol. 1, American Mathematical Society (1999) 41–97.


*V.S. Varadarajan, Supersymmetry for Mathematicians: An Introduction, Courant Lecture Notes 11, 2004.
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$^1$ How the Grassmann-odd indeterminates $\theta$, $\theta_1$, $\theta_2$ are actually defined within the theory of supermanifolds is a question in its own right, cf. e.g. Refs. 1 & 2. In this answer, we will just need some of their basic properties.
A: The Grassmann integral of a gaussian measure with covariance $M^{-1}$ is nothing but the Pfaffian of $M$:
$$
\int \mathrm d\theta_1\mathrm d\theta_2\cdots\mathrm d\theta_{2n}\ \mathrm e^{-\frac12{\vec\theta}{}^tM\vec\theta}\equiv\mathrm{Pf}(M)
$$
Therefore, the question can be recast as: "what is the geometric meaning of the Pfaffian?".
In general terms, the Pfaffian does not really have any deep geometric meaning.  Due to $\mathrm{Pf}(M)^2=\det(M)$, one may think of the Pfaffian as the square root of the volume determined by the columns of $M$. In a more general context, the Pfaffian can be used to define the Euler class of a bundle, but this does not really help in developing a geometric sense of what it does.
More on this: Does the Pfaffian have a geometric meaning?, from Math.OF.
Similarly, integration over Grassmann variables is the exact same thing as differentiation:
$$
\int\mathrm d\theta\ f(\theta)\equiv \frac{\partial}{\partial\theta}f(\theta)
$$
One may therefore also state the question as "what is the geometric meaning of Berezin differentiation". This partially explains the "weird rule" $\mathrm d(a\theta)=a^{-1}\mathrm d\theta$. It also seems to suggest a possible geometric interpretation: as is the case with any derivative, $\partial/\partial\theta$ just provides with the linear approximation to $f(\theta)$ (whatever that means in the Grassmann-valued case). This is not really a meaningful statement, because $f(\theta)$ is already linear to begin with (assuming it is analytic; otherwise, integration/differentiation with respect to $\theta$ is undefined). More generally, integration over $\theta_1\cdots\theta_n$ of a multi-variate Grassmann-valued function just picks up the different coefficients of the Taylor expansion thereof. But this should be thought of as an algebraic exercise rather than a geometric one: here the "polynomial" approximations provided by the Taylor expansion have no pictorial representation in terms of a surface embedded in $\mathbb R^n$. Furthermore, the "error terms" generated by neglecting higher orders are not real-valued, so there is no notion of "smallness".
In short: Grassmann integration is mostly  an algebraic concept, not a geometrical one. The discussion above is admittedly not very illuminating, but it is hard to imagine anything more explicit than that. After all, Grassmann numbers are not a field (not even a ring!), so they cannot be used to define vector spaces (or modules). This seems to preclude any direct geometrical interpretation of these "numbers".
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For fun: one may think of the Faddeev-Popov ghosts of a non-abelian gauge theory as a connection field on a Grassmann-valued fibre over the so-called superspace. When doing so, the Slavnov differential is $s\equiv\partial/\partial\theta$, so that BRST transformations are nothing but translation in the $\theta$ direction. BRST transformations, and their nilpotency, almost acquire a geometrical meaning. The ghost fields are on the same level as the regular gauge fields: they are just a connection over superspace. BRST invariance is just the statement that the physics are invariant under translations in the Grassmann-valued direction, and the nilpotency of the BRST transformations is a straightforward consequence of $\theta^2=0$. The path-integral, being defined as the integral of the gaussian measure over all fields, is just the integral over both $A^\mu$ and $c,\bar c$. Grassmann integration is, algebraically speaking, not different from regular integration, so that all fields are treated symmetrically. Unbeknownst to them, Becchi, Rouet, Stora and Tyutin were in fact the first people to introduce a supersymmetric transformation.
