It quickly becomes apparent, by looking at many accounts of the "manifold assumption," that many writers are notably sloppy about its meaning. The more careful ones define it with a subtle but hugely important caveat: that the data lie on or close to a low-dimensional manifold.
Even those who do not include the "or close to" clause clearly adopt the manifold assumption as an approximate fiction, convenient for performing mathematical analysis, because their applications must contemplate deviations between the data and the estimated manifold. Indeed, many writers later introduce an explicit mechanism for deviations, such as contemplating regression of $y$ against $\mathrm x$ where $\mathrm x$ is constrained to lie on a manifold $M^k\subset \mathbb{R}^d$ but the $y$ may include random deviations. This is equivalent to supposing that the tuples $(\mathrm x_i, y_i)$ lie close to, but not necessarily on, an immersed $k$-dimensional manifold of the form
$$(\mathrm x,f(x)) \in M^k \times \mathbb{R} \subset \mathbb{R}^d\times \mathbb{R}\approx \mathbb{R}^{d+1}$$
for some smooth (regression) function $f:\mathbb{R}^d\to \mathbb{R}$. Since we may view all the perturbed points $(\mathrm x,y)=(\mathrm x,f(\mathrm x)+\varepsilon)$, which are merely close to the graph of $f$ (a $k$ dimensional manifold), as lying on the $k+1$-dimensional manifold $M^k\times \mathbb R$, this helps explain why such sloppiness about distinguishing "on" from "close to" may be unimportant in theory.
The difference between "on" and "close to" is hugely important for applications. "Close to" allows that the data may deviate from the manifold. As such, if you choose to estimate that manifold, then the typical amount of deviation between the data and the manifold can be quantified. One fitted manifold will be better than another when the typical amount of deviation is less, ceteris paribus.
The figure shows two versions of the manifold assumption for the data (large blue dots): the black manifold is relatively simple (requiring only four parameters to describe) but only comes "close to" the data, while the red dotted manifold fits the data perfectly but is complicated (17 parameters are needed).
As in all such problems, there is a tradeoff between the complexity of describing the manifold and the goodness of fit (the overfitting problem). It is always the case that a one-dimensional manifold can be found to fit any finite amount of data in $\mathbb{R}^d$ perfectly (as with the red dotted manifold in the figure, just run a smooth curve through all the points, in any order: almost surely it will not intersect itself, but if it does, perturb the curve in the neighborhood of any such intersection to eliminate it). At the other extreme, if only a limited class of manifolds is allowed (such as straight Euclidean hyperplanes only), then a good fit may be impossible, regardless of the dimensions, and the typical deviation between data and the fit may be large.
This leads to a straightforward, practical way to assess the manifold assumption: if the model/predictor/classifier developed from the manifold assumption works acceptably well, then the assumption was justified. Thus, the appropriate conditions sought in the question will be that some relevant measure of goodness of fit be acceptably small. (What measure? It depends on the problem and is tantamount to selecting a loss function.)
It is possible that manifolds of different dimension (with different kinds of constraints on their curvature) may fit the data--and predict held-out data--equally well. Nothing can be "proven" about "the underlying" manifold in general, especially when working with large, messy, human datasets. All we can usually hope for is that it the fitted manifold is a good model.
If you do not come up with a good model/predictor/classifier, then either the manifold assumption is invalid, you are assuming manifolds of too small a dimension, or you haven't looked hard enough or well enough.
Best Answer
It is indeed an assumption that the function $f$ is "well behaving".
If the function $f:\mathbb R^d \to \mathbb R^m$ were allowed to be arbitrary, then its image $f[\mathbb R^d] \subset \mathbb R^m$ in the target space could have any number of dimensions between $0$ and $m$. The image can e.g. be the whole target space $\mathbb R^m$ (dimensionality $m$), a single point (dimensionality zero), any weird subspace with fractional fractal dimensionality, or whatever.
But if the function $f$ is smooth (i.e. it is continuous and has continuous derivatives of all orders; mathematicians write $f \in C^\infty$), then the image of $f$ will be a differentiable manifold of dimensionality $d$. I think this assumption is implicit in the figure you provided, because the image of $f$ is displayed there as a nice curly surface which is obviously supposed to be "smooth".
Perhaps it is enough that $f$ is continuously differentiable (i.e. $f \in C^1$), which is a weaker requirement (as it assumes nothing about higher derivatives).