Expectation from Joint Distribution We have 
$$f_{X,Y}(x,y) = 
    \begin{cases}
      1 & 0 < x < 1,  x < y < x+1; \\
      0 & \text{otherwise}
    \end{cases}
$$
and we're looking for $\operatorname{Var}Y$.
I got two different answers with two different methods and I'm not sure which one is correct.

Method 1:
$$\operatorname E(Y) = \int_0^1 \int_x^{x+1}yf_{X,Y}(x,y)\,dy\,dx = \int_0^1 \left(x+\frac{1}{2}\right) \, dx = 1$$
$$\operatorname E(Y^2) = \int_0^1 \int_x^{x+1}y^2f_{X,Y}(x,y) \, dy \, dx = \int_0^1 \left(x^2 + x + \frac{1}{3}\right) \, dx = \frac{7}{6}$$
So $\operatorname{Var}(Y) = \frac{1}{6}$

Method 2:
Marginal distribution $f_Y(y) =1$
$$\operatorname E(Y) = \int_x^{x+1}yf_Y(y)\,dy = x+\frac{1}{2}$$
$$\operatorname E(Y^2) = \int_x^{x+1}y^2f_Y(y) \, dy = x^2 + x +\frac{1}{3}$$
So $\operatorname{Var}(Y) = \left(x^2 + x +\frac{1}{3}\right) - \left(x+\frac{1}{2} \right)^2 = \frac{1}{12}$

One of these solutions is probably wrong, but I can't really identify the error.

The first method is correct.
The second method uses the wrong marginal distribution.   It appears to be a joint 
Notice that the joint distribution is uniform over a rhombus.  Plot this and notice the lengths of the horizontal cross section for various $y$ values from $0$ to $2$.
It should be as follows.
$$\begin{split}f_Y(y) &=\int_0^1 \mathbf 1_{x\leq y\leq x+1}\mathrm d x\\&= \int_{\max(0,y-1)}^{\min(1,y)}\mathbf 1_{0\leq y\leq 2}\mathrm d x \\ & = y\mathbf 1_{0\leq y\lt 1}+(2-y)\mathbf 1_{1\leq y\leq 2}\\&=\begin{cases}y &:& 0\leq y\lt 1\\2-y&:&1\leq y\leq 2\\0&:&\text{else}\end{cases} \end{split}$$
As a reality check, a marginal distribution's support should not include other variables, and the interval must contain all realisable values for the variable.
$$\begin{split}1 &=\int_0^1 y\mathrm d y+\int_1^2 (2-y)\mathrm d y\end{split}$$
Then the expectations shall be in agreement with the first method. $$\begin{split}\mathsf E(Y) &=\int_0^1 y^2\mathrm d y+\int_1^2y(2-y)\mathrm d y \\ &= 1\\\mathsf E(Y^2) &=\int_0^1 y^3\mathrm d y+\int_1^2y^2(2-y)\mathrm d y \\ &= \dfrac 76\end{split}$$
