Does the envelope theorem hold at a corner solution?
First, you made a sign error in the computations. After correcting for your error, a crucial hypothesis you miss is that $X$, the choice set, does not depend on the variable $t$ in the theorem (with the notations of the theorem). To apply the theorem properly, the interval $[0,L]$ should not depend on $L$.

A) The sign error

$$\frac{\partial H}{\partial L_K}=-\alpha (L-L_K+1)^{-1}H+(1-\alpha)(L_K+K)^{-1}H=0$$
We define $L_K^0= (1-\alpha)(L+1)-\alpha K$.

B) Why we could think the envelop theorem's result may fail

Presuming that $0&lt;\alpha&lt;1$, there are four possible cases.

(1) $L_K^0\in [0,L]$. One can check the objective function is concave, thus $L_K^*=L_K^0$.

(2.i) $L_K^0\notin [0,L]$ and $H(L,0,K)&lt; H(L,L,K)$. Then $L_K^*=0$.

(2.ii) $L_K^0\notin [0,L]$ and $H(L,0,K)&gt; H(L,L,K)$. Then $L_K^*=L$. 

(2.iii) (just to be exhaustive) $L_K^0\notin [0,L]$ and $H(L,0,K)= H(L,L,K)$. Then there are two solutions, $0$ and $L$. 

In case (1), $$\frac{\partial F}{\partial L}(L,K)=\frac{\partial H}{\partial L}(L,L^*_K,K)+\frac{\partial L^*_K}{\partial L}.\frac{\partial H}{\partial L_K}(L,L^*_K,K).$$
The second term of the right-hand side is equal to zero thanks to the first-order condition. This is compatible with the envelop theorem's result for an interior solution.

In case (2.i), $F(L,K)=H(L,0,K)$ and so $$\frac{\partial F}{\partial L}(L,K)=\frac{\partial H}{\partial L}(L,0,K).$$
This is compatible with the envelop theorem's result for a corner solution here.

In case (2.ii), $F(L,K)=H(L,L,K)$ and so $$\frac{\partial F}{\partial L}(L,K)=\frac{\partial H}{\partial L}(L,L_K=L,K)+\frac{\partial H}{\partial L_K}(L,L_K=L,K).$$

We have to be cautious about the notations here, $\frac{\partial H}{\partial L}$ means the partial derivative corresponding to the first argument, and $\frac{\partial H}{\partial L_K}$ to the second one. The second term of the right-hand side is nonzero, which does not fit with the envelop theorem's result.

C) Why it actually does not fail

Write the problem as $F(L,K)=\max_{x\in [0,1]}H(x,L,K)$, with
$$H(x,L,K)=(L-x L+1)^\alpha(x L+K)^{1-\alpha}.$$
This problem is equivalent to the initial one.
The key difference is that the interval $[0,1]$ does not depend on $L$ or $K$. This is the reason why we can apply the envelop theorem, whereas it was wrong to apply it before. 

We can check that the case (2.ii) is compatible with the envelop theorem, we have $F(L,K)=H(x=1,L,K)$ and so $$\frac{\partial F}{\partial L}(L,K)=\frac{\partial H}{\partial L}(x=1,L,K).$$