Define the operator $\frac{D}{D\tau}$ by its action on an arbitrary contravariant vector $A^\lambda$:
$$\frac{DA^\lambda}{D\tau} = \frac{dA^\lambda}{d\tau} + \Gamma^\lambda_{\mu\nu} \frac{dx^\mu}{d\tau} A^\nu$$
(with the motivation that this allows us to express the geodesic equation in a nice form).
Now, in an attempt to deduce the corresponding action on a covariant vector $\frac{DA_\lambda}{D\tau}$, contract with an arbitrary covariant vector $B_\lambda$:
$$\begin{align} B_\lambda\frac{DA^\lambda}{D\tau} &= B_\lambda\frac{dA^\lambda}{d\tau} + B_\lambda\Gamma^\lambda_{\mu\nu} \frac{dx^\mu}{d\tau} A^\nu \\ &= \frac{d(A^\lambda B_\lambda)}{d\tau} - A^\lambda\frac{dB_\lambda}{d\tau} + B_\nu\Gamma^\nu_{\mu\lambda} \frac{dx^\mu}{d\tau} A^\lambda \\ &= \frac{d(A^\lambda B_\lambda)}{d\tau} - A^\lambda\left(\frac{dB_\lambda}{d\tau} - B_\nu\Gamma^\nu_{\mu\lambda} \frac{dx^\mu}{d\tau}\right) \\ \end{align}$$
The following is the bit which I don't understand - we claim that the term in brackets must be the derivative of this covariant vector, i.e.
$$ \frac{DB_\lambda}{D\tau} = \frac{dB_\lambda}{d\tau} - B_\nu\Gamma^\nu_{\mu\lambda} \frac{dx^\mu}{d\tau} $$
I agree that it is definitely a vector, but I don't see how we can make the leap to saying that it is certainly the form that this operator takes when acting on a covariant vector. Is there something I'm missing?
I considered that it might have something to do with the fact that if you substitute this definition and rearrange you obtain
$$ B_\lambda\frac{DA^\lambda}{D\tau} + A^\lambda\frac{DB_\lambda}{D\tau} = \frac{d(A^\lambda B_\lambda)}{d\tau} $$
but I can't see exactly where this leads.