This doesn't quite answer your question, but it might be useful.

First of all, note that it is extremely unbelievable (please, correct me if I am wrong) that you can glue objects of (bounded) derived category, if you cannot glue morphisms. One reason is that you usually construct gluing "by induction on number of open subsets". But you can't do this induction, if you can't glue morphisms.

Also, from category theoretic point of view it is more reasonable to ask for a functor $R:GluingData(\{U_i\}_{i\in I}) \to D^b(X)$ from the category of gluing data (in bounded derived category) with respect to an open covering $\{U_i\}_{i\in I}$ to the bounded derived category $D^b(X)$.

With all these said, let me show that you can't glue morphisms in (bounded) derived categories. Namely, I will show that morphisms are not uniquely defined by gluing data.

Pick any noetherian scheme X with a non-split short exact sequence of vector bundles
$$
0 \to \mathcal E' \to \mathcal E \to \mathcal E'' \to 0.
$$
[An explicit example of this will be a scheme $\mathbf P^1_k$ and an extension
$$
0 \to \mathcal O(-2) \to \mathcal O(-1)^{\oplus 2} \to \mathcal O \to 0.]
$$
This extension defines a class $\xi \in Ext^1(\mathcal E'', \mathcal E')=Hom_{D^b(X)}(\mathcal E'', \mathcal E'[1])$. So, we can think of $\xi$ as a morphism in the derived category. Now choose any cover of $X$ by open affine subschemes $\{U_i\}_{i\in I}$. The restriction of $\xi$ onto each open $U_i$ is equal to $0$ because we don't have any higher Ext groups between locally free sheaves on affine schemes. Thus $\xi|_{U_i}=0|_{U_i}$ for each $i$, but $\xi \neq 0$. So there is no chance to glue morphisms in (bounded) derived categories.

**UPDATE:** Actually, it seems that you can glue two objects of bounded derived categories. In order to make everything precise let me say what I mean by "bounded derived category". For me $D^b(X)$ will actually mean $D^b(Qcoh(X))$ and I will denote by $D^b_{coh}(X)$ the bounded derived category of coherent sheaves.

**Please, check it carefully. There might be a mistake. The result looks quite mysterious for me.**

**Claim:** Let $X$ be any noetherian scheme and let $U,V$ be two open subschemes of $X$. Given two objects $K\in D^b_{coh}(U), L \in D^b_{coh}(V)$ and an isomorphism $c:K|_{U\cap V} \to L|_{U\cap V}$ we can construct an element $F\in D^b_{coh}(X)$ s.t. $F|_{U}\cong K$ and $F|_{V} \cong L$.

**Proof:** Let us denote the intesection $U\cap V$ by $W$ and consider the derived pushforward functors
$$\mathbf Rj_{U}:D^b(U) \to D^b(X), \mathbf Rj_{V}:D^b(V) \to D^b(X)\text{ and }\mathbf Rj_{W}:D^b(W) \to D^b(X).
$$

Note that there is a natural morphism
$$
b:\mathbf Rj_{V}(L) \to \mathbf Rj_{W}(L|_W)
$$
that is adjoint to the identity morphism
$$
\mathbf Rj_{V}(L)|_{W} \xrightarrow{\mathbf 1} L|_W.
$$
Also, there is a natural morphism
$$
a:\mathbf Rj_{U}(K) \to \mathbf Rj_{W}(L|_W)
$$
that is adjoint to the composition
$$
\mathbf Rj_{U}(K)|_{W} \xrightarrow{\mathbf 1} K|_{W} \xrightarrow{c} L|_{W}.
$$
Taking sum of these morphisms we obtain a map
$$
\phi:\mathbf Rj_{U}(K)\oplus \mathbf Rj_{V}(L) \xrightarrow{(a,b)} \mathbf Rj_{W}(L|_W).
$$
By the **TR3** axiom of triangulated category we can find an object $F\in D^b(X)$ such that the following triangle is distinguished
$$
F\xrightarrow{(f_1,f_2)} \mathbf Rj_{U}(K)\oplus \mathbf Rj_{V}(L) \to \mathbf Rj_{W}(L|_W) \to F[1] (*).
$$
I claim that $F|_{U}$ is isomorphic to $K$ (and the same for $V$ and $L$). Indeed, restrict the triangle $(*)$ on $V$:
$$
F|_{U}\to K\oplus \mathbf Rj_{V}(L)|_{U} \xrightarrow{a|_{U}, b|_{U}} \mathbf Rj_{W}(L|_W)|_{U} \to F|_{U}[1] (*').
$$
But $b|_{U}:\mathbf Rj_{V}(L)|_{U} \to Rj_{W}(L|_W)|_{U}$ is an isomorphism (basically, by definition of $b$). Thus the triangle $(*')$ has a section. The general result about triangulated categories implies that $f_1|_{U}: F|_{U} \to K$ is an isomorphism.

An argument is a bit trickier for $V$ because we a priori know only that $a|_{W}$ is an isomorphism, not $a|_{V}$. In other words, the natural map
$$
f_2|_{V}:F|_{V} \to L
$$
is an isomorphism on $W$. Hence, the support of its cone is concentrated on $Y:=V-W$. Therefore there is a scheme structure on $Y$ s.t. $Cone(a|_{V})=\mathbf Ri_Y(Q)$ for some element $Q\in D^b(Y)$ (where $i_{Y}:Y\to V$ is the natural closed immersion).

In order to show that $f_2|_{V}$ is actually an isomorphism, let us recall Mayer–Vietoris distinguished triangle
$$
F \to \mathbf Rj_{U}((F)|_{U})\oplus \mathbf Rj_{V}((F)|_{V}) \to \mathbf Rj_{W}((F)|_{W}) \to F[1] (**).
$$

Observe that there is a morphism of distinguished triangles induced by a pair $(\mathbf Rj_{U}(f_1|_{U}), \mathbf Rj_{V}(f_2|_{V}))$ (I don't know how to type commutative diagrams here)
$$
F \to \mathbf Rj_{U}((F)|_{U})\oplus \mathbf Rj_{V}((F)|_{V}) \to \mathbf Rj_{W}((F)|_{W}) \to F[1] (**) \\
\downarrow \\
F\to \mathbf Rj_{U}(K)\oplus \mathbf Rj_{V}(L) \to \mathbf Rj_{W}(L|_W) \to F[1] (*).
$$
The first map $F \to F$ is an isomorphism by definition, the third map $\mathbf R j_W F|_W \to \mathbf R j_W L|_{W}$ is an isomorphism since $f_1|_{W}$ and $f_2|_{W}$ are isomorphisms. Thus we conclude that the second morphism
$$
\mathbf Rj_{U}((F)|_{U})\oplus \mathbf Rj_{V}((F)|_{V}) \to \mathbf Rj_{U}(K)\oplus \mathbf Rj_{V}(L)
$$
is an isomorphism. In particular, $\mathbf Rj_{V}(f_2|_{V})$ is an isomorphism. Apply this result to a distinguished triangle
$$
F|_{V} \xrightarrow{f_2|_{V}} L \to \mathbf Ri_{Y}(Q) \to F|_{V}[1]
$$
to obtain the result that
$$
0\cong \mathbf Rj_{V} \mathbf Ri_{Y}(Q) \cong \mathbf Ri_{Y\to X}Q.
$$
On the other hand, $\mathbf Li_{Y\to X}\mathbf Ri_{Y\to X} Q \cong Q$ since $i_{Y \to X}$ is a finite map. Thus we see that
$$
Q\cong \mathbf Li_{Y\to X}\mathbf Ri_{Y\to X} Q \cong 0.
$$
So, the cone $Cone(f_2|_{V}) \cong 0$. In other words, $f_2|_{V}$ is an isomorphism. Therefore, we see that $K|_{V} \cong L$.

The last thing to check is that $K\in D^b_{Coh}(X)$, but it is clear since $K|_{U}\in D^b_{Coh}(U)$ and $K|_{V}\in D^b_{Coh}(V)$. So, we are done!

choiceof an isomorphism $P_1 \vert_{U_1 \cap U_2} \cong P_2 \vert_{U_1 \cap U_2}$ defines a unique sheaf $P$ on $U_1 \cup U_2$. Different choices of isomorphism will lead to different $P$'s. $\endgroup$ – Dan Petersen Nov 28 '17 at 7:14