Please see this diagram. Lines $m,n$ are normals to reflexive axes with the angle between them $\frac\theta2$. Then $v''$, which is reflected twice by $m,n$ is such a vector rotated $\theta$ from the original vector $v$.
And I think this has also an algebraic explanation in geometric algebra. The reflection of $v$ by the axis $n$ is represented as $v'=-nvn$. Then $v''=-mv'm=-m(-nvn)m=(mn)v(nm)=RvR^\dagger$, where $R=mn$ and $R^\dagger$ is reverse of $R$. We speak of $R$ is rotor of angle $\theta$ if $m\cdot n=\cos\frac\theta2$. $RvR^\dagger$ is exactly the expression of a rotation in geometric algebra.
Write $c := \cos\theta$, $s := \sin\theta$, $\mathbf{w} := \mathbf{u}\times\mathbf{v} = s\mathbf{n}$, and $T := T_\mathbf{u}\left(T_\mathbf{v}\right)$, so that we have ...
$$\begin{align}
T(\mathbf{x}) &=\mathbf{x}-2(\mathbf{x}\cdot\mathbf{u})\mathbf{u}-2(\mathbf{x}\cdot\mathbf{v})\mathbf{v} + 4c(\mathbf{x}\cdot\mathbf{v})\mathbf{u}\\
R(\mathbf{x}) &= (2c^2-1)\mathbf{x} + 2 s^2 (\mathbf{x}\cdot\mathbf{n}) \mathbf{n} + 2 s c (\mathbf{x}\times\mathbf{n}) \\
&= (2c^2-1)\mathbf{x} + 2 (\mathbf{x}\cdot\mathbf{w})\mathbf{w} + 2 c (\mathbf{x}\times\mathbf{w}) \\
\end{align}$$
Decomposing $\mathbf{x}$ as $p\mathbf{u} + q\mathbf{v} + r \mathbf{w}$, we can get fairly directly ...
$$\mathbf{x}\cdot\mathbf{u} = p + q c \qquad \mathbf{x}\cdot\mathbf{v}=pc+q \qquad \mathbf{x}\cdot\mathbf{w}=rs^2 \qquad (\star)$$
$$\mathbf{x}\times\mathbf{w} = \mathbf{x}\times \left(\mathbf{u}\times\mathbf{v}\right) = (\mathbf{x}\cdot\mathbf{v})\mathbf{u}-(\mathbf{x}\cdot\mathbf{u})\mathbf{v} \qquad (\star\star)$$
Then it's straightforward to show that the difference of the transformations vanishes:
$$\begin{align}
T(\mathbf{x}) - R(\mathbf{x}) &=\mathbf{x}-2(p+qc)\mathbf{u}-2(pc+q)\mathbf{v} + 4c(pc+q)\mathbf{u}\\
&-\left( (2c^2-1)\mathbf{x} + 2 r s^2 \mathbf{w} + 2 c \left( (pc+q)\mathbf{u} - (p+qc)\mathbf{v} \right) \right) \\[6pt]
&= (2-2c^2)\;\mathbf{x} + 2 \left(-p-qc+2pc^2+2qc-pc^2-qc\right)\;\mathbf{u} \\
&+ 2\left(-pc-q+pc+qc^2\right)\mathbf{v} - 2 r s^2 \mathbf{w} \\[6pt]
&= 2 s^2 \left( \mathbf{x} - p\mathbf{u} - q \mathbf{v} - r\mathbf{w} \right) \\[6pt]
&= 0
\end{align}$$
and we conclude that the transformations are equivalent. $\square$
Edit. Without jumping immediately to the decomposition of $\mathbf{x}$, we can use the expansion in $(\star\star)$ to write
$$\begin{align}
\frac{T(\mathbf{x})-R(\mathbf{x})}{2s^2} \;\;&=\;\; \mathbf{x}
\;-\; \left( \; \frac{\mathbf{x}.( \mathbf{u} - c \mathbf{v} )}{s^2}\;\mathbf{u}
\;+\; \frac{\mathbf{x}.( \mathbf{v} - c \mathbf{u} )}{s^2}\;\mathbf{v}
\;+\; \frac{\mathbf{x}.\mathbf{w}}{s^2}\;\mathbf{w} \;\right)
\end{align}$$
If you can "see" that the coefficients of $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ are the components of $\mathbf{x}$ ---which would be clear for orthogonal $\mathbf{u}$ and $\mathbf{v}$, for which $c=0$ and $s=1$--- then you're done. If not, note that you can arrive at this insight by solving the dot-product equations $(\star)$ for $p$, $q$, $r$.
Best Answer
If you're working with a real inner product on $V$ with positive definite signature then we can describe the product of two reflections geometrically.
First, some linear algebra. Say $V$ is an internal direct sum of subspaces, $V=V_1\oplus V_2$. Suppose further that $A$ is a linear map on $V$ which preserves $V_1$ and $V_2$. That is, it restricts to linear maps $A_1$ and $A_2$ on them. Then we may use the notation $A=A_1\oplus A_2$. (Conversely, given linear maps $A_1$ and $A_2$ on $V_1$ and $V_2$, there is a unique linear map $A$ on $V$ that restricts to them.) If we can do the same thing for another linear map $B$ on $V$, then we have $A+B=(A_1+B_1)\oplus (A_2+B_2)$ and $AB=(A_1B_1)\oplus(A_2B_2)$. In other words, there is an algebra homomorphism
$$ \mathrm{End}(V_1)\times\mathrm{End}(V_2)\to \mathrm{End}(V_1\oplus V_2). $$
Thus, to compose $A$ and $B$ we may find subspaces they stabilize and compose the two transformations "componentwise." This works with more than two components too.
Suppose $A$ is a reflection on $V$ across a hyperplane $\Pi$ perpendicular to some one-dimensional subspace $P$. Then $A$ stabilizes $P$ and $\Pi$. Moreoever, it acts as the identity on $\Pi$ and the negation map on $P$. Then suppose $B$ is another reflection, this time across a hyperplane perpendicular to a different one-dimensional subspace $L$. Now let's write $V=P\oplus L\oplus W$ where $W$ is the orthogonal complement of $\mathrm{span}(P,L)=P+L$. Then $A$ and $B$ act as $\mathrm{id}$ on $W$, so $AB$ acts as $\mathrm{id}$ on $W$ as well, and it suffices to see how $AB$ acts on the two-dimensional subspace $P+L$.
For this, we can draw a picture:
This plane represents the span of $P$ and $L$. The axis $P'$ and $L'$ are the lines perpendicualr to $P$ and $L$ respectively. Starting with a vector $x$, flip it across $P'$ to get $y$, then flip $y$ across $L'$ to get $z$. The effect of going from $x$ to $z$ is moving by an angle of $2(a+b)$. But that's exactly twice the angle between $P'$ and $L'$, which is independent of the original vector $x$, so all vectors are rotated by an angle of $2(a+b)$.
The orthogonal group $O(V)$ has two connected components. The connected component of the identity is $SO(V)$, the group of rotations. The other connected component includes hyperplane reflections, but it also generally includes other transformations which are not hyperplane reflections (such as $-I_3\in O(3)$, which has no nonzero fixed points, let alone a hyperplane's worth). In any case, these two connected components are precisely the fibers of $\det:O(V)\to\{\pm1\}$. That is, an element of $O(V)$ is a rotation if it has determinant $1$ and has determinant $-1$ otherwise.