I think that the answer to the question is basically that one usually considers rectifiable sets in the sense you give and further specifies that they are measurable.

On the other hand, I do not think you technically need measurability for a.e. tangent planes. Recall that your definition of rectifiability is equivalent to demanding that up to a set of measure zero, the set is contained in a countable union of $C^1$-manifolds. One way to get at the tangent plane of a rectifiable set is to say that a point $x \in S$ has a tangent plane $T_xM_i$ when $x\in M_i$ as long as for $x\in M_j$ for some other $j$ has $T_xM_j= T_xM_i$. Then one would have to check that this is defined fo $H^m$-a.e. for $x\in S$, but this follows from a sort of transversality argument, intuitively because $M_i$ and $M_j$ cannot intersect very often if they do not have the same tangent planes at their intersections. So, this argument should go through without reference to the measurability of $S$ at all.

This said, as far as I know, people work with rectifiable sets which are also measurable, so it is just an issue of symantics whether or not it is included in the definition.

Rectifiable (measurable) sets can still be very bad though. For example, if you take circles inside $B(0,1)\subset \mathbb{R}^2$, which are centered at the rational points and have radii which are square summable, then the union of these is a $1$-rectifiable set, because it is obviously a countable union of (smooth) $1$-manifolds. However it is dense in the ball! On the other hand, it has tangent planes almost everywhere!

So, rectifiability does not really avoid "square filling things" in this sense.

If it helps, an example of a non-rectifiable set is as follows. Take a triangle. Remove a regular hexagon inscribed in the triangle (side length 1/3 that of the triangle). This gives three triangles. Continue doing this and then take the intersection of all of these. This is a purely $1$-rectifiable set, roughly because if you project on to an axis perpendicular to a base of the triangle, then you get a cantor set, of $H^1$-measure zero and it is not hard to see that a $1$-rectifiable set must have at most one direction with this property (this set has three).

This is in response to your comment on your answer:

You ask why your original definition and this one are the same. Let me list three equivalent definitions of $k$-rectifiability:

(1) Up to a set of measure zero $S \subset \cup_{i=1}^\infty M_i$ where the $M_i$ are $C^1$ $k$-dimensional submanifolds.

(2) Up to a set of measure zero $S \subset \cup_{i=1}^\infty F_i(\mathbb{R}^k)$ for $F:\mathbb{R}^k\to \mathbb{R}^N$ (where $\mathbb{R}^N$ is the ambient space).

(3) Up to a set of measure zero $S\subset \cup_{i=1}^\infty \text{image}(F_i)$ where the $F_i :T_i \to \mathbb{R}^N$ is $C^1$ and $T_i$ is a $k$-plane in $\mathbb{R}^N$ where if we denote $\Pi_{T_i}$ by the projection $\mathbb{R}^N\to T_i$, we have that $\Pi_{T_i} \circ F_i = Id$.

The equivalence of (1) and (2) is in Leon Simon's book, I believe. It uses a.e. differentiability of Lipschitz functions and a Sard type theorem. To understand (3), notice that the bit about the projection just says that the images of $F_i$ are just graphs over the $T_i$. Graphs are $C^1$ submanifolds, so this (3) => (1). On the other hand any $C^1$ manifold is locally the graph over its tangent plane, so we can slice up a decomposition as in (1) into countably many such graphs, implying (3).