No, not necessarily.

Consider the case $B=kH$, $C=kG$ of finite group algebras over a field $k$, where $H\leq G$. I'll write $\downarrow$ and $\uparrow$ for restriction and induction. This case has a couple of simplifying features. Firstly, induction is both left and right adjoint to restriction. Secondly, $kH$ is selfinjective, and so the projective module $U_i$ is also injective. If the answer to the question were "yes", then there would be an injection from $U_i$ to a direct sum of copies of $V_{ij}\!\downarrow$. But since $U_i$ is injective, this would split, and so, by Krull-Schmidt, $U_i$ would be a direct summand of $V_{ij}\!\downarrow$.

Suppose that $U_i$ is the projective cover of the simple $kH$-module $S$ and $V_{ij}$ the projective cover of the simple $kG$-module $T$.

Then $V_{ij}$ is a direct summand of $U_i\!\uparrow$ if and only if
$$\text{Hom}_{kG}(U_i\!\uparrow, T)\cong\text{Hom}_{kH}(U_i,T\!\downarrow)$$
is nonzero, which is the case if and only if $T\!\downarrow$ has a composition factor isomorphic to $S$.

$U_i$ does *not* occur as a direct summand of $V_{ij}\!\downarrow$ if and only if
$$\text{Hom}_{kH}(V_{ij}\!\downarrow,S)\cong\text{Hom}_{kG}(V_{ij},S\!\uparrow)$$
*is* zero, which is the case if and only if $S\!\uparrow$ does *not* have a composition factor isomorphic to $T$.

So we just need to find simple modules $S$ for $kH$ and $T$ for $kG$ for which $S$ is a composition factor of $T\!\downarrow$ but $T$ is not a composition factor of $S\!\uparrow$.

This happens quite commonly, but $S$ can't appear in the head or socle of $T\!\downarrow$, or else there would be a map in one direction between $S$ and $T\!\downarrow$, and hence between $S\!\uparrow$ and $T$, and so $T$ would appear in the socle or head of $S\!\uparrow$. So $T$ has to be reasonably large.

One example that I happen to be familiar with is where $k$ is algebraically closed of characteristic two, $G=A_5$ and $H=A_4$. Then $kG$ has a $4$-dimensional simple projective module $T$ and $kH$ has three one-dimensional simple modules: take $S$ to be one of the non-trivial ones. Then $S$ and $T$ have the required properties.