Why, intuitively, is the order reversed when taking the transpose of the product?

One of my best college math professor always said:

Make a drawing first.

Although, he couldn't have made this one on the blackboard.

By dualizing $AB: V_1\stackrel{B}{\longrightarrow} V_2\stackrel{A}{\longrightarrow}V_3$, we have $(AB)^T: V_3^*\stackrel{A^T}{\longrightarrow}V_2^*\stackrel{B^T}{\longrightarrow}V_1^*$.

Edit: $V^*$ is the dual space $\text{Hom}(V, \mathbb{F})$, the vector space of linear transformations from $V$ to its ground field, and if $A: V_1\to V_2$ is a linear transformation, then $A^T: V_2^*\to V_1^*$ is its dual defined by $A^T(f)=f\circ A$. By abuse of notation, if $A$ is the matrix representation with respect to bases $\mathcal{B}_1$ of $V_1$ and $\mathcal{B}_2$ of $V_2$, then $A^T$ is the matrix representation of the dual map with respect to the dual bases $\mathcal{B}_1^*$ and $\mathcal{B}_2^*$.

Here's another argument. First note that if $v$ is a column vector then $(Mv)^T = v^T M^T$. This is not hard to see - if you write down an example and do it both ways, you will see you are just doing the same computation with a different notation. Multiplying the column vector $v$ on the right by the rows of $M$ is the same as multiplying the row vector $v^T$ on the left by the columns of $M^T$.

Now let $( \cdot , \cdot )$ be the usual inner product on $\mathbb{R}^n$, that is, the dot product. Then the transpose $N = M^T$ of a matrix $M$ is the unique matrix $N$ with the property

$$(Mu, v) = (u, Nv).$$

This is just a consequence of associativity of matrix multiplication. The dot product of vectors $u,v$ is given by thinking of $u,v$ as column vectors, taking the transpose of one and doing the dot product: $(u,v) = u^T v$.

Then $(Mu,v) = (Mu)^T v = (u^T M^T) v = u^T (M^Tv) = (u, M^Tv)$.

Exercise: Show uniqueness!

With this alternate definition we can give a shoes-and-socks argument. We have

$$( ABu, v) = (Bu, A^Tv) = (u, B^TA^Tv)$$

for all $u,v$, and so $(AB)^T = B^T A^T$. The argument is exactly the same as the one for inverses, except we are "moving across the inner product" instead of "undoing".