Is this adaptation of one-time-pad still secure? Having shared a pad of length double the length of your anticipated message, splitting it into x and y.

Then using x to encrypt the message, and using y twice to encrypt the next pad, would that be insecure (seeing as the data I'm encrypting would still be random)?
No, it's not secure; after three messages, the attacker can gain information about the second and third messages.

To review this in greater detail, lets look at your proposal, and what it actually exposes to the user:

To start with, you have a secret $x_0$ and $y_0$ (I'll subscript them to distinguish the values between the iterations).  To encrypt the first message $m_0$, you generate a fresh secret $x_1$, $y_1$, and transmit:

$$m_0 \oplus x_0, x_1 \oplus y_0, y_1 \oplus y_0$$

To send the second message $m_1$, you generate more secrets $x_2$, $y_2$, and transmit:

$$m_1 \oplus x_1, x_2 \oplus y_1, y_2 \oplus y_1$$

For the third message $m_2$, you generate even more secrets $x_3$, $y_3$, and transmit:

$$m_2 \oplus x_2, x_3 \oplus y_2, y_3 \oplus y_2$$

At this point, the attacker has seen $x_1 \oplus y_0, y_1 \oplus y_0, m_1 \oplus x_1, x_2 \oplus y_1, m_2 \oplus x_2$.  If xor's all those values together, he gets:

$$x_1 \oplus y_0 \oplus y_1 \oplus y_0 \oplus m_1 \oplus x_1 \oplus x_2 \oplus y_1\oplus m_2 \oplus x_2 = m_1 \oplus m_2$$

And thus he has recovered the difference between the two messages.  And, other than the fact that the attack cannot recover the first message (because the attacker doesn't get any information on $x_0$), this is not really any better than xoring in a constant string.