Stéphane is right to say that $Y$ cannot stay undefined, you have to give a value to $Y(\omega)$ for all $\omega\in\Omega$.
Stéphane’s choice of $Y = \infty$ when $X \le d$ is a bit difficult to handle even if it is natural. Let’s take $Y = y_0\in\mathbb R$ when $X \le d$. You have,
$$\begin{align*}
\mathbb P( Y \le y ) &= \mathbb P( y_0 \le y \cap X \le d \cup X-d \le y \cap X > d ) \\
&= \mathbb P( y_0 \le y \cap X \le d) + \mathbb P(d < X \le y+d) \\
&= \mathbb P(X\le d) \cdot 1_{y_0\le y} + \left\{\begin{array}{l}
0 \text{ if } y < 0 \\
F(y+d)-F(d) \text{ if } y \ge 0.
\end{array}\right.
\end{align*}$$
This cdf has no derivative in $y = y_0$, hence strictly speaking no density.
However defining $g(y)$ by
$$g_0(y) =\left\{\begin{array}{l}
0 \text{ if } y < 0 \\
f(y+d) \text{ if } y \ge 0,
\end{array}\right. $$
(the derivative of the pdf "ignoring" the step in $y_0$) and $p = \mathbb P(X>d) = 1 - F(d)$, we can write the density of $Y$ by the mixture
$$g(y) = g_0(y) + (1-p) \cdot \delta(y-y0)$$
where $\delta$ is the Dirac function.
Note that the integral $\int_{-\infty}^\infty g_0(y) \mathrm dy$ is equal to $p$ and by convention the integral of the $\delta$ function is 1, so $\int_{-\infty}^\infty g(y) \mathrm dy = 1$.
Edit 1 you can chose $y_0 = +\infty$ is you wish, in that case there is a Dirac mass in $+\infty$. I don’t know if there is some standard notation for this.
Edit 2 I suspect that the original question was to give the density of $Y$ conditional to $X>d$.
In that case you don’t need bother with the values taken by $Y$ when this condition is not fulfilled. You can write
$$\begin{array}{rcl}
\mathbb P( Y \le y | X > d) &=& { \mathbb P(X \le y+d \cap X > d) \over \mathbb P(X>d)} \\
&=& \left\{\begin{array}{l}
0 \text{ if } y < 0 \\
{F(y+d)-F(d)\over 1 - F(d)} \text{ if } y \ge 0,
\end{array}\right.
\end{array}$$
and the density of $Y$ conditional to $X>d$ is
$$g(y|X>d) = \left\{\begin{array}{l}
0 \text{ if } y < 0 \\
{1\over 1-p}f(y+d) \text{ if } y \ge 0.
\end{array}\right. $$
As whuber pointed out in the comments, a natural interpretation of this is to change the sample space $\Omega$ to $\Omega' = \{ \omega \in \Omega \>:\> X(\omega) > d \}$.
